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1. <! --<TO-DETERMINE>-->
Normal force acting on the billiard ball at point B
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<! --<ANSWER>-->
1.143BF N
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Given info:
Billiard ball of,
Radius= 0 2.50r cm
Mass= 160.0om g
Translational speed of billiard ball at A = 1
2.00ms
0v
Point B is at the top of a hill that has a radius of curvature of 60cm ( )r
Billiard ball rolls without slipping down the track.
Formula used:
Let’s name normal force acting on the ball at point B
As BF .
Let’s name the angular velocity of billiard ball at the top of the track as .
Let’s name the linear speed of billiard ball at the top of the hill as sv
Let’s name the moment of inertia of billiard ball as sI .
Let’s name the vertical distance from potential energy zero level h 10cm
2
10g ms
.
Conservation of mechanical energy:
i i f fK U K U
2. Kinetic energy for an object that undergoes both translation and rotation:
2 21 1
2 2
s sK Mv I ...(1)
Condition for rolling without slipping:
s ov r ...(2)
Centrifugal power of billiard ball,
2
s
B o
v
F m
r
Calculation:
Let’s consider the motion of sphere,
Initially the billiard ball is at rest with translational kinetic energy, so 2
0 0
1
2
iK m v .
The initial gravitational potential energy is (0) 0i oU m g
Final gravitational potential energy is ( )f oU m g h
Conservation of mechanical energy:
i i f fK U K U
2
0 0
1
(0) ( )
2
f om v K m g h
21
2
f o o oK m v m gh ...( )A
Let’s consider the kinetic energy fK of billiard ball at the top of the hill,
Kinetic energy is part translational and part rotational. We can use (2) equation to write
In terms of sv .
Using (1) expression,
Kinetic energy for an object that undergoes both translation and rotation:
3. 2 21 1
2 2
s sK Mv I ...(1)
2 21 1
2 2
f o s sK m v I
Condition for rolling without slipping:
s ov r ...(2)
s
o
v
r
Substitute into kinetic energy equation:
2 21 1
2 2
f o c cK m v I
2
21 1
2 2
s
f o s s
o
v
K m v I
r
2
2
0
1
2
s
f o s
I
K m v
r
From the general knowledge we know that moment of inertia of a sphere is 2
0
2
5
om r .
So, let’s substitute the sI value in to the equation,
2
2
0
1
2
s
f o s
I
K m v
r
2
0
2
2
2
1 5
2
o
f o s
o
m r
K m v
r
2
0
1 2
2 5
f o sK m m v
2
0
1 7
2 5
f sK m v
27
10
f o sK m v ...( )B
4. Since ( ),( )A B equations are equal,
( ) ( )A B
2 2
0 0
1 7
( )
2 10
o o sm v m g h m v
2 2
0
1 7
( )
2 10
sv g h v
2 2
0
10 1
( )
7 2
sv g h v
...(3)
Let’s consider the billiard ball at the top of the hill,
Centrifugal power of billiard ball,
2
s
B o
v
F m
r
Substituting 2
sv value to above equation,
2
s
B o
v
F m
r
2
0
10 1
( )
7 2
B o
v g h
F m
r
2
010 2 ( )
14
B o
v g h
F m
r
Let’s substitute the values,
2
010 2 ( )
14
B o
v g h
F m
r
1 2 2 2
10 (2.00 ) 2*10 *( 10*10 )
14
B o
ms ms m
F m
r
2 2 2 2
10 (4.00 2.00
14
B o
m s m s
F m
r
3 2 2
2
160*10 *10 6.00
14*60*10
B
kg m s
F
m