A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed
//tex.z-dn.net/?f=v_%7Bcom%7D" id="TexFormula1" title="v_{com}" alt="v_{com}" align="absmiddle" class="latex-formula">,0 = 9.0 m/s and initial angular speed ω₀ = 0. The coefficient of kinetic friction between the ball and the lane is 0.23. The kinetic frictional force f with 
acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed 
has decreased enough and angular speed ω has increased enough, the ball stops sliding and then rolls smoothly.
(a) What then is
in terms of ω? m·ω
(b) During the sliding, what is the ball's linear acceleration? m/s²
(c) During the sliding, what is the ball's angular acceleration? rad/s²
(d) How long does the ball slide? s
(e) How far does the ball slide? m
(f) What is the speed of the ball when smooth rolling begins? m/s
(a) What then is
(b) During the sliding, what is the ball's linear acceleration? m/s²
(c) During the sliding, what is the ball's angular acceleration? rad/s²
(d) How long does the ball slide? s
(e) How far does the ball slide? m
(f) What is the speed of the ball when smooth rolling begins? m/s
1 answer:
Answer:
a) v_com= Rω
b) -2.254 m/
c) 51.2 rad/
d) t=1.08 seconds
e) x=7.865m
f) v_roll=6.07m
Explanation:
Initially, the ball is travelling with v_com=v_0
Wen not rotating, at the initial stage the ball must be sliding along the surface.
This motion therefore generates a frictional force F_r at the point of contact.
Let the velocity at the point of contact be v_bottom
v_bottom=v_com-Rω
Therefore when ω=0, v_bottom=v_com
So when the ball begins rolling
v_com= Rω
F_r=μ_rmg
〖-F〗_r=ma_com
a_com=(〖-μ〗_r mg)/m
a_com=-μ_rg
a_com=-(0.23)(9.8)
a_com=-2.254m/s^2
Te negative sow decrearse
=(μ_r mgR)/I = (〖5μ〗_r mgR)/2mRR
=(〖5μ〗_r g)/2R
=(5*(0.23)*(9.8))/(2*0.11)
=51.2 rad/s^2
t=v_0/(〖-a〗_com+Rα)
=8.5/(2.255+0.11*(51.2))
=8.5/7.886
=1.08 seconds
X=v_0 t+1/2 a_com t^2
X=8.5*(2.254) - 1/2 (2.254)*〖1.08〗^2
=7.865m
v_roll=v_0+a_com t_r
=8.5-(2.254)(1.08)
=6.07m/sec
Initially, the ball is travelling with v_com=v_0
Wen not rotating, at the initial stage the ball must be sliding along the surface.
This motion therefore generates a frictional force F_r at the point of contact.
a) Let the velocity at the point of contact be v_bottom
v_bottom=v_com-Rω
Therefore when ω=0, v_bottom=v_com
So when the ball begins rolling
v_com= Rω
b) F_r=μ_rmg
〖-F〗_r=ma_com
a_com=(〖-μ〗_r mg)/m
a_com=-μ_rg
a_com=-(0.23)(9.8)
a_com=-2.254m/s^2
Te negative sow decrearse
c) α=(μ_r mgR)/I = (〖5μ〗_r mgR)/2mRR
=(〖5μ〗_r g)/2R
=(5*(0.23)*(9.8))/(2*0.11)
=51.2 rad/s^2
d) t=v_0/(〖-a〗_com+Rα)
=8.5/(2.255+0.11*(51.2))
=8.5/7.886
=1.08 seconds
e) X=v_0 t+1/2 a_com t^2
X=8.5*(2.254) - 1/2 (2.254)*〖1.08〗^2
=7.865m
f) v_roll=v_0+a_com t_r
=8.5-(2.254)(1.08)
=6.07m/sec
You might be interested in
First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:
1) Take first force, say
, and assume that it is pointing towards the x-direction.
Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
= 
, where 
= Unit vector in the x-direction.
2) Take the second force, say
, and assume that it is making an angle 
with the first force .
Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:
Now from simple vector addition, we know that,
--- (A)
Where
= Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.
Let us find
first!
= 
=> = 
Now the magnitude of is,
|| = 
=> || = 
Since
, therefore,
=> || = 
Since || = ||, and the magnitude of the resultant force is 20N, therefore,
|| = | |
20 =
Take square on both sides,
400 =
Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°
-israr
1) Take first force, say
Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
2) Take the second force, say
Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:
Now from simple vector addition, we know that,
Where
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.
Let us find
=>
Now the magnitude of
|
=> |
Since
=> |
Since |
|
20 =
Take square on both sides,
400 =
Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°
-israr
Answer: Work Done would remain same.
Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity . If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height
. If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.
We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.