ALP Sol P Circular Motion E

8/18/2019 ALP Sol P Circular Motion E

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SOLN_Circular Motion - 1

TOPIC : CIRCULAR MOTION

PART - I

1. at = 0 = constant

2. Velocity chnages as its direction changeAcceleration changes as its direction change.

3. |v| = constant

4. T = m!2RT + 60 = m (3!)2R

m!2

R = 8

60

!2R = 2.5 m/s2

9!2R = 22.5 m/s2

5. (C) as # = 0 $ at = 0

% FC = ma

C = m!2R = constant

6. mg = constant vector

7. When critical velocities are taken T = 6mgmax

A mg

TB

BTD

D

T = 0A

C

TA + T

C = 6 mg = constant

TB + T

D = 6 mg = constant

8. Let v be the speed of particle at B, just when it is about to loose contact.From application of Newton's second law to the particle normal to the spherical surface.

r

mv 2

= mg sin & .......... (1)

Applying conservation of energy as the block moves from A to B..

2

1 mv2 = mg (r cos # – r sin &) .......... (2)

Solving 1 and 2 we get3 sin & = 2 cos #

9. % T2 # d3 $

3

13

122

2

1

10

10

T

T

'

'

(

)

*

*

+

, -

'

'

(

)

*

*

+

, $

2

1

T

T =

1010

1

% T1 : T

2 = 1 : 1010

10. Initial extension will be equal to 6 m.

% Initial energy =2

1 (200) (6)2 = 3600 J.

Reaching A :2

1mv2 = 3600 J

$ mv2 = 7200 J

From F.B.D. at A :

N =5

7200

R

mv 2- = 1440 N

PHYSICS SOLUTIONS OF"ADVANCED LEVEL PROBLEMS"

Target : ISEET (IIT-JEE)




11. Since F!

. V!

, the particle will move along a circle.

% F =R

mv 2

& / =R

S$ / = 2mv

FS

12. x2 = 4ayDifferentiating w.r.t. y, we get

dx

dy

= a2

x

% At (2a, a),dx

dy = 1 $ hence / = 45°

the component of weight along tangential direction is mg sin /.

hence tangential acceleration is g sin / =2

g

13. As tangential acceleration a = dV/dt = !dr/dt

but ! = 20n= 40 rad/s anddt

dr = (2 × 0.75) × 10 –2 m/s = 1.5 × 10 –2 m/s (reel is turned uniformly at the rate

of 2 r.p.s.)

% a = 40 × 1.5 × 10 –2 m/s2 = 60 × 10 –2 m/s2

Now by the F.B.D. of the mass.

w

T

T – W = ag

W% T = W (1 + a/g) put a = 60 × 10 –2 % T = 1.019 W

14. For A :

T – m!2r – ma = 0 .............(i)Seen from object itself

T – 3

mg = ma .............(i)

For B :mg – T = ma .............(ii)(i) – (ii)

2T =3

4mg

T =3

2mg

15. |v| = constant

16. ! = constant

% # = 0 = constant




17. F = 22t

2 )mg(Ff 11 > mg#

f

F

mg

(f = m r)!2

(asdt

Vd = constant % F

t = constant)

Now when the angular speed of the rod isincreasing at const. rate the resultant force

will be more inclined towards f! .

Hence the angle between F!

and horizontal plane decreases so as with the rod due to increases in

f = m!2r only.

PART - II

1. (a) The system is in equilibrium whenm

1gsin/ = m

2g

or sin/ =1

2

m

m

(b) Let the tangential acceleration of m1 be a.% m

2g – m

1g sin/ = (m

1 + m

2) a

a =9

2540 2 =

9

15m/s2

th e normal acceleration of m1 is zero. # speed of m

1 is zero.

% The magnitude of acceleration of m1 =

9

15m/s2 .

2. (a) Applying conservation of energy between initial and final position isLoss in gravitational P.E. of the bead of mass m = gain in spring P. E.

% mg R =

2

1 K (2R – 2 R)2

or K =)223(R

mg

2(b) At t = 0a

t = g

ac = 0

at lowest pointa

t = 0

ac = 0

The centripetal acceleration of bead at the initial and final position is zero because its speed at both positionis zero.The tangential acceleration of the bead at initial position is g.

The tangential acceleration of the bead at lower most position is zero.

3. Since belt is not slipping, speed at rim of A and B is same

!Ar

A = !

Br

B

!A = 100 ×

10

25 = 250 rpm = 250 ×

60

20 rad/sec.

=3

250 rad/sec.

! = !0 + #t

t = 2 /

03

25

0

20

= 3

50

sec.




4. !0

2 = 900 (rad/sec)2 $ !0 = 30 rad/sec.

!2 = 1600 (rad/sec)2 $ ! = 40 rad/sec.

/ = t2

0 ' (

)*+

, !1!

t =3040

21002

1033

=7

400sec.

5. (a) As a rod AB moves, the point ‘P’ will always lie on the circle.

% its velocity will be along the circle as shown by ‘VP’ in the figure.

If the point P has to lie on the rod ‘ AB’ also then it should have

component in ‘x’ direction as ‘V’.

% VP sin / = V $ V

P = V cosec /

here cos/ =R

x =

R

1 .

5

R3 =

5

3

% sin/ =5

4% cosec / =

4

5

% VP = 45 V ...Ans.

(b) ! =R

VP =

R4

V5

ALTERNATIVE SOLUTION :(a) Let ‘P’ have coordinate (x, y)

x = R cos /, y = R sin /.

VX =

dt

dx = – R sin /

dt

d/ = V $

dt

d/ =

/2sinR

V

and VY = R cos /

dtd/ = R cos / ''

( )**

+ , /2 sinR

V = – V cot /

% VP = 2

y2x VV 1 = /1 222 cotVV = V cosec / ...Ans.

(b) ! =R

VP =

R4

V5

6. Taking a small element at DN sin/ = dmg

N =/sin

dmg

2T sin4 –

N cos/ = dm!2

R2T sin4 = dm(!2R + g cot/)




But 4 is very small, sin4 5 4

2T 4 =R2

md

0$

(!2R + g cot/)

' (

)*+

, R2

dT2 $

=R2

md

0$

(!2R + g cot/)

T = '' (

)

**+

,

/1

!

0 cotg

R

2

mg 2

.

7. The particle velocity has two components.(i) v

0 sin# vertical which move it in vertical direction.

(ii) v0 cos# in horizontal direction and along the cylindrical surface which cause it to move in circle.

So, N =R

)cosv(m 20 #

=R

cosmv 220 #

8. amF!!

- or )aa(mF yx

!!!

1- (# a2 = 0)

x = a sin !t

vx

=dt

dx = aw cos (!t)

ax = 2

2

dt

xd = – a!2 sin(!t)

vy =

dt

dy = – b !sin(!t)

ay = 2

2

dt

yd = – b !2 cos(!t)

So ) j tcosbitsina(mF 22 !!2!!2-!

) jˆ tcosbi

ˆtsina(mF 2 !1!!2-

!

) j yix(mF 2 1!2-!

222 yxm|F| 1!-!

direction tan# =x

y =

a

bcot(!t) (from x-axis)

or )] jyix[( 1 is position vector of the particle in corrdinate system. Because of negative sign force is

opposite to it and always acting towards the orzon.




9. In x-direction,

ax = g sin# –

g sin# cos6ax = g sin# (1 – cos6)

vx = g sin# 7 62

t

0

dt)cos1( .........(i)

In tangential direction, As a rod AB moves, the point ‘P’ will always lie on the circle.

at = g sin# cos6 – 8g cos#

at = –g sin# (1 – cos6)

v – v0 = – gsin# 7 62

t

0

dt)cos1( .........(ii)

from (i) & (ii)v – v

0 = –v

x = – v cos6

v =)cos1(

v0

61 .

10. Net tangential force acting on the element due to gravity is

d mg sin/ gS

Total external force on chain along the length is

F = 7 /sindmg

F = //7 dRsingm

R /

0

$

$ $ a = 7 //-R /

0

dsin gR

m

F $

$

a = 9 : R / 0cos

gR $

$/2 $ a = ;

<

=>?

@' (

)*+

, 2

Rcos1

gR $

$.

11. !t = a

So, v = at = as2

also, !N =

R

v2

bt4 =R

ta 22

t2 =bR

a2

and bt4 =R

as2

22

bR

a b '

' (

)**+

, =

R

as2

R =bs2

a3

$ ! = 2N

2t !1! = (# !

N=

R

v2

=R

ta 22

= 2

4

bR

a)

! =

2

2

42

bR

a

a '' (

)

**+

,

1 $ ! =

2

3

2

a

bS4

1a ''

(

)

**

+

,

1 .




12. (a) Parabola y = ax2 is shown. It is clear from diagram that at x = 0 velocity is along x-axis and constant aN

is along y-axis. So,

aN = 2

2

dt

yd

dt

dy = 2a ×

dt

dx = 2aVx

2

2

dt

yd = 2av

dt

dx = 2av2 (# 0

dt

xd2

2

- )

aN = 2av2

R = 2

2

av2

v =

a2

1.

(b) 1b

y

a

x2

2

2

-' (

)*+

, 1'

(

)*+

,

Here again at x = 0 particle is at (0, ± b) moving along positive or negative x-axis hence aN is along y-axis

only.

aN = 2

2

dt

yd

0dt

dy

b

y2

dt

dx

a

x222

-1

0dt

dy

b

y2

a

vx222 -'

(

)*+

, 1

0dt

yd

b

y2

dt

dy

b

2

dt

dx

a

v22

2

2

2

22 -'

' (

)**+

, 1'

(

)*+

, 1 [# v = const. along x-axis only

dt

dy = 0]

'' (

)**+

, 2-

2

2

22

2

dt

yd

b

)b(2

a

v2a

N = 2

2

a

bv2 R =

N

2

a

v2 =

b

a2

13. at =

dt

dv = a = 0.50 m/s2

at = #R

/ =2t

2

1#

$ s = R/ =

2

1 #Rt2 $ s =

2

1 a

tt2

(0.1) 20R =2

1 (0.5)t2 $ t =

5

R40

v = a . t = 0.55

R40

ac =

R

v2

=5

425.0 03 = 0.2 0

a = 2t

2t aa 1 = A B A B2

5

2

21 01 =

2510

41 1 =

2013 m/s2




14. V = 10 m/s

tan / =Rg

v2

$ / = tan –1 º451010

1010-'

(

)*+

, 33

15. (i) At # = 0º

A

T

mg

anet

= CR

v2

acceleration is vertically up(ii) At # = 90º is at B

v = 0

mg

Acceleration is vertically down.(iii) Horizontally

#||

|

|

atotal

mga = gsint #

a =C

v2

R

u=0

#

R cos #

tan # = #sing

R / v2

$ g sin # . tan # =R

v2

.....(1)

Using energy conservation :

#- cosmgRmv2

1 2....(2)

By (1) & (2)

tan # = 2

1

% cos # = 3

1

% # = cos –1 '' (

)**+

,

3

1

16. (a) N1 =

R

mv2

=5010

1001

33

8N

mg

N1

CN1 = 0.2 N

(b) N =R

mv2

cos / ....(1)

for just slipping

8N =R

mv2

sin / .....(2)

from eqn (1) & (2)

tan / = 8 =3

1=

58.0

1 = 1.724 $ / = 30º Ans.

17. a =12

12

mm

mm

12

!2R =3

R2!

T =12

21

mm

mm2

1 !2R =3

4 m!2R




18. / =4

0

tan 45º =c

t

a

a$ a

t = a

C

$ g sin # =

t

v2

...(1)

#

$ $ – cos #

3g$

v=?

Using energy conservation

2

1 m 3g$ –

2

1 mv2 = mg $ (1 – cos #)

$ mv2 = 3mg$ – 2mg$ + 2mg$ cos #mv2 = mg$ + 2mg$ cos # .............(2)

By eq. (1) and (2)sin # = 1 + 2 cos # $ # = 90º

19.mgcosN

RmsinN 2

-/!-/

tan / =g

R2!

$hr

R

2 =

g

R2!$ !2 = g/(r – h) ......(1)

(a) h > 0 $ r – 2

g

! > 0

$ ! > r / g =1.0

8.9 = 98 = 27 rad/second Ans.

(b) g = !2 (r – h)

h

h

g

g D2-

D = –

h

10 42

maximum value of h is 0.1 so that Dg = –h

10g 423 = – 9.8 × 10 –3 m/s2 Ans.

20. N – mg sin 30º = m!2R .....(1)

30ºN

!

mg cos30º

8N

mg sin30º

mg cos 30º = 8N .....(2)

! = 20 rad/s

$2

mg3 = 8 ;

<

=>?

@!1 Rm

2

mg 2

$ R)3(2

g 2!-82

R = 22

)3(g

!

82

R = 2)2(2

6.03(8.9

0

2 = 0.24 m Ans.

For minimum angular velocity, normal sould be zero at heighest pointm!2 R = mg

! = R

g

= 24.0

8.9

= 6.4 rad/second




Also, condition for which block will not slip on cylinder isN – mg cos/ = m!2RN = mg cos/ + m!2Rfr max

= 8N = 8(mg cos/ + m!2R)For the block does not slip over cylinder,

mg sin/ E frmax

mg sin/ E 8 mg cos/ + 8 m!2R

R

cosmg –sing

8

//F!

)cos(sinR

g/82/

8F!

block will not shift anywhere if ! is greater than maximum possible value of RHS which is

2 / 12 )1(R

g81

8F!

! F 8.9 rad/sec.

!min

= 8.9 rad/sec.

21. # = kt

a

ac

/ atat = #r = ktr

# =dt

d! = kt $ 77 -!

! t

00

ktdtd

! =2

kt2

, aC = !2r = r

4

tk 42

tan/ =t

c

a

a =

ktr

4 / rtk 42

=4

kt3

$ t =

3 / 1

k

tan4' (

)*+

, /

22. Centripetal acceleration at A = !2R

acceleration along AB = !

2

R cos /Time taken to reach the point B

L = 0 +2

1 (!2 R cos /)t2 (L << R)

t =/! cosR

L22 Ans.

23. T = kx = 147 (0.1 sec / – 0.1)

T sin / = m !2 r

$ 147(0.1 sec / – 0.1)sin / = 0.3 × (14)2 (0.1 tan /)

1 – cos / = 0.4

$ cos/ =5

3$ / = 53º

T = 147 (0.1 sec 53 – 0.1) = 9.8 N

N = T cos/ – mg = 9.8 ×5

3 – 0.3 × 9.8 = 2.94 N

N = 2.94 N Ans.




TOPIC : CIRCULAR MOTION

PART - I

1. at = 0 = constant

2. Velocity chnages as its direction changeAcceleration changes as its direction change.

3. |v| = constant

4. T = m!2RT + 60 = m (3!)2R

m!2

R = 8

60

!2R = 2.5 m/s2

9!2R = 22.5 m/s2

5. (C) as # = 0 $ at = 0

% FC = ma

C = m!2R = constant

6. mg = constant vector

7. When critical velocities are taken T = 6mgmax

A mg

TB

BTD

D

T = 0A

C

TA + T

C = 6 mg = constant

TB + T

D = 6 mg = constant

8. Let v be the speed of particle at B, just when it is about to loose contact.From application of Newton's second law to the particle normal to the spherical surface.

r

mv 2

= mg sin & .......... (1)

Applying conservation of energy as the block moves from A to B..

2

1 mv2 = mg (r cos # – r sin &) .......... (2)

Solving 1 and 2 we get3 sin & = 2 cos #

9. % T2 # d3 $

3

13

122

2

1

10

10

T

T

'

'

(

)

*

*

+

, -

'

'

(

)

*

*

+

, $

2

1

T

T =

1010

1

% T1 : T

2 = 1 : 1010

10. Initial extension will be equal to 6 m.

% Initial energy =2

1 (200) (6)2 = 3600 J.

Reaching A :2

1mv2 = 3600 J

$ mv2 = 7200 J

From F.B.D. at A :

N =5

7200

R

mv 2- = 1440 N

PHYSICS SOLUTIONS OF"ADVANCED LEVEL PROBLEMS"

Target : ISEET (IIT-JEE)




11. Since F!

. V!

, the particle will move along a circle.

% F =R

mv 2

& / =R

S$ / = 2mv

FS

12. x2 = 4ayDifferentiating w.r.t. y, we get

dx

dy

= a2

x

% At (2a, a),dx

dy = 1 $ hence / = 45°

the component of weight along tangential direction is mg sin /.

hence tangential acceleration is g sin / =2

g

13. As tangential acceleration a = dV/dt = !dr/dt

but ! = 20n= 40 rad/s anddt

dr = (2 × 0.75) × 10 –2 m/s = 1.5 × 10 –2 m/s (reel is turned uniformly at the rate

of 2 r.p.s.)

% a = 40 × 1.5 × 10 –2 m/s2 = 60 × 10 –2 m/s2

Now by the F.B.D. of the mass.

w

T

T – W = ag

W% T = W (1 + a/g) put a = 60 × 10 –2 % T = 1.019 W

14. For A :

T – m!2r – ma = 0 .............(i)Seen from object itself

T – 3

mg = ma .............(i)

For B :mg – T = ma .............(ii)(i) – (ii)

2T =3

4mg

T =3

2mg

15. |v| = constant

16. ! = constant

% # = 0 = constant




17. F = 22t

2 )mg(Ff 11 > mg#

f

F

mg

(f = m r)!2

(asdt

Vd = constant % F

t = constant)

Now when the angular speed of the rod isincreasing at const. rate the resultant force

will be more inclined towards f! .

Hence the angle between F!

and horizontal plane decreases so as with the rod due to increases in

f = m!2r only.

PART - II

1. (a) The system is in equilibrium whenm

1gsin/ = m

2g

or sin/ =1

2

m

m

(b) Let the tangential acceleration of m1 be a.% m

2g – m

1g sin/ = (m

1 + m

2) a

a =9

2540 2 =

9

15m/s2

th e normal acceleration of m1 is zero. # speed of m

1 is zero.

% The magnitude of acceleration of m1 =

9

15m/s2 .

2. (a) Applying conservation of energy between initial and final position isLoss in gravitational P.E. of the bead of mass m = gain in spring P. E.

% mg R =

2

1 K (2R – 2 R)2

or K =)223(R

mg

2(b) At t = 0a

t = g

ac = 0

at lowest pointa

t = 0

ac = 0

The centripetal acceleration of bead at the initial and final position is zero because its speed at both positionis zero.The tangential acceleration of the bead at initial position is g.

The tangential acceleration of the bead at lower most position is zero.

3. Since belt is not slipping, speed at rim of A and B is same

!Ar

A = !

Br

B

!A = 100 ×

10

25 = 250 rpm = 250 ×

60

20 rad/sec.

=3

250 rad/sec.

! = !0 + #t

t = 2 /

03

25

0

20

= 3

50

sec.




4. !0

2 = 900 (rad/sec)2 $ !0 = 30 rad/sec.

!2 = 1600 (rad/sec)2 $ ! = 40 rad/sec.

/ = t2

0 ' (

)*+

, !1!

t =3040

21002

1033

=7

400sec.

5. (a) As a rod AB moves, the point ‘P’ will always lie on the circle.

% its velocity will be along the circle as shown by ‘VP’ in the figure.

If the point P has to lie on the rod ‘ AB’ also then it should have

component in ‘x’ direction as ‘V’.

% VP sin / = V $ V

P = V cosec /

here cos/ =R

x =

R

1 .

5

R3 =

5

3

% sin/ =5

4% cosec / =

4

5

% VP = 45 V ...Ans.

(b) ! =R

VP =

R4

V5

ALTERNATIVE SOLUTION :(a) Let ‘P’ have coordinate (x, y)

x = R cos /, y = R sin /.

VX =

dt

dx = – R sin /

dt

d/ = V $

dt

d/ =

/2sinR

V

and VY = R cos /

dtd/ = R cos / ''

( )**

+ , /2 sinR

V = – V cot /

% VP = 2

y2x VV 1 = /1 222 cotVV = V cosec / ...Ans.

(b) ! =R

VP =

R4

V5

6. Taking a small element at DN sin/ = dmg

N =/sin

dmg

2T sin4 –

N cos/ = dm!2

R2T sin4 = dm(!2R + g cot/)




But 4 is very small, sin4 5 4

2T 4 =R2

md

0$

(!2R + g cot/)

' (

)*+

, R2

dT2 $

=R2

md

0$

(!2R + g cot/)

T = '' (

)

**+

,

/1

!

0 cotg

R

2

mg 2

.

7. The particle velocity has two components.(i) v

0 sin# vertical which move it in vertical direction.

(ii) v0 cos# in horizontal direction and along the cylindrical surface which cause it to move in circle.

So, N =R

)cosv(m 20 #

=R

cosmv 220 #

8. amF!!

- or )aa(mF yx

!!!

1- (# a2 = 0)

x = a sin !t

vx

=dt

dx = aw cos (!t)

ax = 2

2

dt

xd = – a!2 sin(!t)

vy =

dt

dy = – b !sin(!t)

ay = 2

2

dt

yd = – b !2 cos(!t)

So ) j tcosbitsina(mF 22 !!2!!2-!

) jˆ tcosbi

ˆtsina(mF 2 !1!!2-

!

) j yix(mF 2 1!2-!

222 yxm|F| 1!-!

direction tan# =x

y =

a

bcot(!t) (from x-axis)

or )] jyix[( 1 is position vector of the particle in corrdinate system. Because of negative sign force is

opposite to it and always acting towards the orzon.




9. In x-direction,

ax = g sin# –

g sin# cos6ax = g sin# (1 – cos6)

vx = g sin# 7 62

t

0

dt)cos1( .........(i)

In tangential direction, As a rod AB moves, the point ‘P’ will always lie on the circle.

at = g sin# cos6 – 8g cos#

at = –g sin# (1 – cos6)

v – v0 = – gsin# 7 62

t

0

dt)cos1( .........(ii)

from (i) & (ii)v – v

0 = –v

x = – v cos6

v =)cos1(

v0

61 .

10. Net tangential force acting on the element due to gravity is

d mg sin/ gS

Total external force on chain along the length is

F = 7 /sindmg

F = //7 dRsingm

R /

0

$

$ $ a = 7 //-R /

0

dsin gR

m

F $

$

a = 9 : R / 0cos

gR $

$/2 $ a = ;

<

=>?

@' (

)*+

, 2

Rcos1

gR $

$.

11. !t = a

So, v = at = as2

also, !N =

R

v2

bt4 =R

ta 22

t2 =bR

a2

and bt4 =R

as2

22

bR

a b '

' (

)**+

, =

R

as2

R =bs2

a3

$ ! = 2N

2t !1! = (# !

N=

R

v2

=R

ta 22

= 2

4

bR

a)

! =

2

2

42

bR

a

a '' (

)

**+

,

1 $ ! =

2

3

2

a

bS4

1a ''

(

)

**

+

,

1 .




12. (a) Parabola y = ax2 is shown. It is clear from diagram that at x = 0 velocity is along x-axis and constant aN

is along y-axis. So,

aN = 2

2

dt

yd

dt

dy = 2a ×

dt

dx = 2aVx

2

2

dt

yd = 2av

dt

dx = 2av2 (# 0

dt

xd2

2

- )

aN = 2av2

R = 2

2

av2

v =

a2

1.

(b) 1b

y

a

x2

2

2

-' (

)*+

, 1'

(

)*+

,

Here again at x = 0 particle is at (0, ± b) moving along positive or negative x-axis hence aN is along y-axis

only.

aN = 2

2

dt

yd

0dt

dy

b

y2

dt

dx

a

x222

-1

0dt

dy

b

y2

a

vx222 -'

(

)*+

, 1

0dt

yd

b

y2

dt

dy

b

2

dt

dx

a

v22

2

2

2

22 -'

' (

)**+

, 1'

(

)*+

, 1 [# v = const. along x-axis only

dt

dy = 0]

'' (

)**+

, 2-

2

2

22

2

dt

yd

b

)b(2

a

v2a

N = 2

2

a

bv2 R =

N

2

a

v2 =

b

a2

13. at =

dt

dv = a = 0.50 m/s2

at = #R

/ =2t

2

1#

$ s = R/ =

2

1 #Rt2 $ s =

2

1 a

tt2

(0.1) 20R =2

1 (0.5)t2 $ t =

5

R40

v = a . t = 0.55

R40

ac =

R

v2

=5

425.0 03 = 0.2 0

a = 2t

2t aa 1 = A B A B2

5

2

21 01 =

2510

41 1 =

2013 m/s2




14. V = 10 m/s

tan / =Rg

v2

$ / = tan –1 º451010

1010-'

(

)*+

, 33

15. (i) At # = 0º

A

T

mg

anet

= CR

v2

acceleration is vertically up(ii) At # = 90º is at B

v = 0

mg

Acceleration is vertically down.(iii) Horizontally

#||

|

|

atotal

mga = gsint #

a =C

v2

R

u=0

#

R cos #

tan # = #sing

R / v2

$ g sin # . tan # =R

v2

.....(1)

Using energy conservation :

#- cosmgRmv2

1 2....(2)

By (1) & (2)

tan # = 2

1

% cos # = 3

1

% # = cos –1 '' (

)**+

,

3

1

16. (a) N1 =

R

mv2

=5010

1001

33

8N

mg

N1

CN1 = 0.2 N

(b) N =R

mv2

cos / ....(1)

for just slipping

8N =R

mv2

sin / .....(2)

from eqn (1) & (2)

tan / = 8 =3

1=

58.0

1 = 1.724 $ / = 30º Ans.

17. a =12

12

mm

mm

12

!2R =3

R2!

T =12

21

mm

mm2

1 !2R =3

4 m!2R




18. / =4

0

tan 45º =c

t

a

a$ a

t = a

C

$ g sin # =

t

v2

...(1)

#

$ $ – cos #

3g$

v=?

Using energy conservation

2

1 m 3g$ –

2

1 mv2 = mg $ (1 – cos #)

$ mv2 = 3mg$ – 2mg$ + 2mg$ cos #mv2 = mg$ + 2mg$ cos # .............(2)

By eq. (1) and (2)sin # = 1 + 2 cos # $ # = 90º

19.mgcosN

RmsinN 2

-/!-/

tan / =g

R2!

$hr

R

2 =

g

R2!$ !2 = g/(r – h) ......(1)

(a) h > 0 $ r – 2

g

! > 0

$ ! > r / g =1.0

8.9 = 98 = 27 rad/second Ans.

(b) g = !2 (r – h)

h

h

g

g D2-

D = –

h

10 42

maximum value of h is 0.1 so that Dg = –h

10g 423 = – 9.8 × 10 –3 m/s2 Ans.

20. N – mg sin 30º = m!2R .....(1)

30ºN

!

mg cos30º

8N

mg sin30º

mg cos 30º = 8N .....(2)

! = 20 rad/s

$2

mg3 = 8 ;

<

=>?

@!1 Rm

2

mg 2

$ R)3(2

g 2!-82

R = 22

)3(g

!

82

R = 2)2(2

6.03(8.9

0

2 = 0.24 m Ans.

For minimum angular velocity, normal sould be zero at heighest pointm!2 R = mg

! = R

g

= 24.0

8.9

= 6.4 rad/second




Also, condition for which block will not slip on cylinder isN – mg cos/ = m!2RN = mg cos/ + m!2Rfr max

= 8N = 8(mg cos/ + m!2R)For the block does not slip over cylinder,

mg sin/ E frmax

mg sin/ E 8 mg cos/ + 8 m!2R

R

cosmg –sing

8

//F!

)cos(sinR

g/82/

8F!

block will not shift anywhere if ! is greater than maximum possible value of RHS which is

2 / 12 )1(R

g81

8F!

! F 8.9 rad/sec.

!min

= 8.9 rad/sec.

21. # = kt

a

ac

/ atat = #r = ktr

# =dt

d! = kt $ 77 -!

! t

00

ktdtd

! =2

kt2

, aC = !2r = r

4

tk 42

tan/ =t

c

a

a =

ktr

4 / rtk 42

=4

kt3

$ t =

3 / 1

k

tan4' (

)*+

, /

22. Centripetal acceleration at A = !2R

acceleration along AB = !

2

R cos /Time taken to reach the point B

L = 0 +2

1 (!2 R cos /)t2 (L << R)

t =/! cosR

L22 Ans.

23. T = kx = 147 (0.1 sec / – 0.1)

T sin / = m !2 r

$ 147(0.1 sec / – 0.1)sin / = 0.3 × (14)2 (0.1 tan /)

1 – cos / = 0.4

$ cos/ =5

3$ / = 53º

T = 147 (0.1 sec 53 – 0.1) = 9.8 N

N = T cos/ – mg = 9.8 ×5

3 – 0.3 × 9.8 = 2.94 N

N = 2.94 N Ans.

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ALP Sol P Circular Motion E