View
217
Download
1
Category
Tags:
Preview:
Citation preview
Acceleration Due to Acceleration Due to GravityGravity
• Every object on the earth Every object on the earth experiences a common force: experiences a common force: the force due to gravity.the force due to gravity.
• This force is always directed This force is always directed toward the center of the toward the center of the earth (downward).earth (downward).
• The acceleration due to The acceleration due to gravity is relatively constant gravity is relatively constant near the Earth’s surface.near the Earth’s surface.
• Every object on the earth Every object on the earth experiences a common force: experiences a common force: the force due to gravity.the force due to gravity.
• This force is always directed This force is always directed toward the center of the toward the center of the earth (downward).earth (downward).
• The acceleration due to The acceleration due to gravity is relatively constant gravity is relatively constant near the Earth’s surface.near the Earth’s surface.
Earth
Wg
Gravitational AccelerationGravitational Acceleration
• In a vacuum, all objects fall In a vacuum, all objects fall with same acceleration.with same acceleration.
• Equations for constant Equations for constant acceleration apply as acceleration apply as usual.usual.
• Near the Earth’s surface:Near the Earth’s surface:
• In a vacuum, all objects fall In a vacuum, all objects fall with same acceleration.with same acceleration.
• Equations for constant Equations for constant acceleration apply as acceleration apply as usual.usual.
• Near the Earth’s surface:Near the Earth’s surface:
aa = g = = g = 9.81 m/s9.81 m/s22
Directed downward (usually Directed downward (usually negative).negative).
Sign Convention:Sign Convention: A Ball Thrown A Ball Thrown
Vertically Vertically UpwardUpward
• Velocity is positive (+) Velocity is positive (+) or negative (-) based or negative (-) based on on direction of motiondirection of motion..
• Velocity is positive (+) Velocity is positive (+) or negative (-) based or negative (-) based on on direction of motiondirection of motion..
• Displacement is positive Displacement is positive (+) or negative (-) (+) or negative (-) based on based on LOCATIONLOCATION. .
• Displacement is positive Displacement is positive (+) or negative (-) (+) or negative (-) based on based on LOCATIONLOCATION. .
Release Point
UP = +
TippensTippens
• Acceleration is (+) or (-) Acceleration is (+) or (-) based on direction of based on direction of forceforce (weight). (weight).
y = 0
y = +
y = +
y = +
y = 0
y = -NegativeNegative
v = +
v = 0
v = -
v = -
v= -NegativeNegative
a = -
a = -
a = -
a = -
a = -a = -
Same Problem Solving Same Problem Solving Strategy Except Strategy Except aa = g = g::
Draw and label sketch of problem.Draw and label sketch of problem.
Indicate Indicate ++ direction and direction and forceforce direction. direction.
List givens and state what is to be found.List givens and state what is to be found.
Given: ____, _____, a = - 9.8 m/s2
Find: ____, _____ Select equation containing one and
not the other of the unknown quantities, and solve for the unknown.
Example 7:Example 7: A ball is thrown vertically A ball is thrown vertically upward with an initial velocity of upward with an initial velocity of 30 m/s30 m/s. . What are its position and velocity after What are its position and velocity after 2 s2 s, , 4 s4 s, and , and 7 s7 s??
Step 1. Draw and label a sketch.
a = g
+
vo = +30 m/s
Step 2. Indicate + direction and force direction.Step 3. Given/find info.
a = -9.8 ft/s2 t = 2, 4, 7 s
vo = + 30 m/s y = ? v = ?
Finding Finding Displacement:Displacement:
a = g
+
vo = 30 m/s
0
y = y = (30 m/s)(30 m/s)tt + + ½½(-9.8 (-9.8 m/sm/s22))tt22
Substitution of t = 2, 4, and Substitution of t = 2, 4, and 7 s will give the following 7 s will give the following values: values:
y = 40.4 m; y = 41.6 m; y = -30.1 my = 40.4 m; y = 41.6 m; y = -30.1 m
210 0 2y y v t at
Step 4. Select equation that contains y and not v.
Finding Velocity:Finding Velocity:Step 5. Find v from equation that contains v and not x:
Step 5. Find v from equation that contains v and not x:
Substitute t = 2, 4, and 7 Substitute t = 2, 4, and 7 s:s:
v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/sv = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s
a = g
+
vo = 30 m/s
0fv v at 0fv v at
230 m/s ( 9.8 m/s )fv t
Example 7: (Cont.)Example 7: (Cont.) Now Now find the find the maximum heightmaximum height attained:attained:
Displacement is a Displacement is a maximum when the maximum when the velocity velocity vvff is zero. is zero. a =
g
+
vo = +96 ft/s
230 m/s ( 9.8 m/s ) 0fv t
2
30 m/s; 3.06 s
9.8 m/st t
To find To find yymaxmax we we substitute substitute tt = 3.06 s = 3.06 s into the general into the general equation for equation for displacement.displacement.y = y = (30 m/s)(30 m/s)tt + + ½½(-9.8 (-9.8
m/sm/s22))tt22
Example 7: (Cont.)Example 7: (Cont.) Finding the Finding the maximum maximum height:height:
y = y = (30 m/s)(30 m/s)tt + + ½½(-9.8 (-9.8 m/sm/s22))tt22
a = g+
vo =+30 m/s
tt = 3.06 = 3.06 ss
212(30)(3.06) ( 9.8)(3.06)y
yy = 91.8 m - 45.9 = 91.8 m - 45.9 mm
Omitting units, we obtain:Omitting units, we obtain:
ymax = 45.9 m
CONCLUSION OF CONCLUSION OF Chapter 6 - AccelerationChapter 6 - Acceleration
Recommended