Mechanics: A bowling ball is thrown and rolls with slipping

In summary: So from the definition of angular momentum:$$L=\frac{mv^2}{2}$$$$L=0$$So the ball doesn't move at all.
  • #1
enthusiast
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Summary:: A bowling ball is thrown on a bowling lane with the coefficient of friction ##\mu## with an initial velocity of ##v_0## and no rotation. After hitting the ground, it starts rolling with slipping. After how much time after hitting the lane will the ball stop slipping? What will be its velocity in the moment, when it stops slipping?

So I wanted to practice some problems on rigid body mechanics and found this one that I have problems with.

A person in a bowling alley throws a ball towards a lane. When thrown, the ball has a forward velocity ##v_0## and does not experience rotation. It hits the ground and start to roll with slipping. The coefficient of friction between the ball and the bowling lane is ##\mu##. After some time the ball slows down and rolls without slipping. The task is to calculate the time it will take the ball to stop slipping and its velocity at the moment of it stopping to slip.

I was able to derive the equation for time, but I have problems with calculating the velocity. Firstly, I will show how I calculated the time.

I started by calculating the force of friction ##F_f=mg\mu##, where ##m## is the ball's mass. The only force affecting the ball is the force of friction, so it is the source of its acceleration (or rather deceleration): ##F_f=ma##, where ##a## is the ball's acceleration. Following:
$$mg\mu=ma$$ $$g\mu=a$$ $$g\mu=\frac {\Delta v} {t}$$ $$t=\frac{\Delta v}{g\mu}$$
Where ##\Delta v## is the change in the ball's velocity. I called the velocity of the ball in the moment of it stopping slipping ##v_1##. This gives:
$$t=\frac{v_0-v_1}{g\mu}$$
Now onto the second part. I tried calculating ##v_1## three times and each time I got a different result.

The first time, I used the conservation of energy. I started with ##E_{k_0}=E_{k_1}+E_{r_1}##, where ##E_{k_0}## is the initial kinetic energy of the ball, ##E_{k_1}## is the kinetic energy of the ball at the moment, it stop slipping and ##E_{r_1}## is its rotational energy at that moment. After expanding I got:
$$\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+\frac{I\omega_1^2}{2}$$
Where ##I## is the ball's inertial moment and ##\omega_1## is the ball's rotational velocity when it stops slipping. I substituted ##I## with ##\frac{2}{5}mr^2##(as the ball is a solid sphere) and ##\omega_1## with ##\frac{v_1}{r}##(as at this moment the ball is no longer slipping) and got:
$$\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+\frac{\frac{2}{5}mr^2*(\frac{v_1}{r})^2}{2}$$ $$mv_0^2=mv_1^2+\frac{2}{5}mr^2*(\frac{v_1}{r})^2$$ $$v_0^2=v_1^2+\frac{2}{5}v_1^2$$ $$v_1^2=\frac{5v_0^2}{7}$$ $$v_1=\sqrt{\frac{5}{7}}v_0$$
This seems close to the correct answer of ##v_1=\frac{5}{7}v_0##, but its quite different. Anyway, then I realized that I forgot to include the work of the force of friction (##W_f##) . The original equation of ##E_{k_0}=E_{k_1}+E_{r_1}## changed to ##E_{k_0}=E_{k_1}+E_{r_1}+W_f##. I started to calculate ##W_f## like ##W_f=F_ts## where ##s## is the distance (spatium from Latin) traveled by the ball. To calculate the distance, I used ##s=\frac{v_0+v_1}{2}*t## and then substituted ##t## with what I got earlier: ##s=\frac{v_0+v_1}{2}*\frac{v_0-v_1}{g\mu}##. I substituted in ##W_f=F_t*s##:
$$W_f=mg\mu*\frac{v_0+v_1}{2}*\frac{v_0-v_1}{g\mu}$$ $$W_f=\frac{m(v_0^2-v_1^2)}{2}$$
After substituting in ##E_{k_0}=E_{k_1}+E_{r_1}+W_f## I got:
$$\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+\frac{I\omega_1^2}{2}+\frac{m(v_0^2-v_1^2)}{2}$$ $$\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+\frac{\frac{2}{5}mr^2*(\frac{v_1}{r})^2}{2}+\frac{m(v_0^2-v_1^2)}{2}$$
$$v_0^2=v_1^2+\frac{2}{5}v_1^2+(v_0^2-v_1^2)$$
$$0=\frac{2}{5}v_1^2$$
Which is nonsense. I have no idea where did I make a mistake and would be thankful for an explanation. I hope I explained my notation well, as on StackExchange Physics everyone seemed offended by me not using, what they called "proper notation".
Thanks.
 
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  • #2
Your energy balance equation does not include the correct energy lost to friction.
The equation
$$\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+\frac{\frac{2}{5}mr^2*(\frac{v_1}{r})^2}{2}+\frac{m(v_0^2-v_1^2)}{2}$$is nonsense because, as you have discovered, it simplifies to $$ \frac{\frac{2}{5}mr^2*(\frac{v_1}{r})^2}{2}=0.$$Like I said, you need to account for friction correctly in terms of ##\mu##.

On edit: Even if you wrote down the energy equation correctly, you need to be exra careful because the mechanical energy lost to friction is not related to the distance traveled by the CM but to the distance along the circumference of the ball over which slipping takes place.

You get the result much more quickly by noting that the only horizontal force acting on the ball is friction and that the torque due to friction about any point lying on the surface is zero. This means that angular momentum about any point lying on the surface is conserved.
 
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  • #3
enthusiast said:
I used the conservation of energy.
Have you tried conservation of momentum (linear & angular)?
 
  • #4
The ball is a rigid body, its centre of mass translates, and the ball rotates about its centre of mass. You need to take both motions in consideration.
The force of Kinetic friction decelerates the translation, but its torgue accelerates rotation. You have equations for both: the speed v decreases while the angular speed w increases up to the moment when the rolling condition v=wr fulfills: From that moment, the ball rolls without slipping.
What are the time functions of the velocity and angular velocity v(t) and w(t)?
 
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  • #5
  • #6
kuruman said:
You get the result much more quickly by noting that the only horizontal force acting on the ball is friction and that the torque due to friction about any point lying on the surface is zero. This means that angular momentum about any point lying on the surface is conserved.
Yes, that is a very neat trick in finding the eventual velocity. But the question also asks for the time taken to achieve rolling contact, so I think there is no way to avoid analysing forces and torques.
 
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  • #7
haruspex said:
Yes, that is a very neat trick in finding the eventual velocity. But the question also asks for the time taken to achieve rolling contact, so I think there is no way to avoid analysing forces and torques.
Yes the time depends on ##\mu## which is brought in through forces and torques. It is useful though to know the final velocity of the CM from other considerations.
 
  • #8
haruspex said:
Yes, that is a very neat trick in finding the eventual velocity. But the question also asks for the time taken to achieve rolling contact, so I think there is no way to avoid analysing forces and torques.

I agree that it's much more straightforward to go via. torques and forces. I suppose if you really wanted to, you could equivalently write the conservation equations and differentiate. The power of the friction force is, if the velocity of the point instantaneously in contact with the ground is ##\vec{v} = \vec{V} + \vec{\omega} \times \vec{r} = (V - \omega r) \hat{x}##,$$P = \vec{F} \cdot \vec{v} = F_x(V-\omega r)$$where ##F_x = -F## is the force of friction. The total energy of the system would be$$E = \frac{1}{2}mV^2 + \frac{1}{2}I_z\omega^2$$ $$\frac{dE}{dt} = MV \dot{V} + I_z\omega \dot{\omega} = -F(V-\omega r)$$The angular momentum equation, about a coordinate system fixed to the floor, is $$\frac{dL_z}{dt} = \frac{d}{dt} (I_z\omega + MVr) = I_z\dot{\omega} + Mr\dot{V} = 0$$Finally the momentum equation is$$F_x = -F = M\dot{V} \implies \dot{V} = -\frac{F}{M} \implies V(t) = v_0 - \frac{F}{M}t$$Substituting into the energy equation gives $$I_z \omega \dot{\omega} = F\omega r \implies \dot{\omega} = \frac{Fr}{I_z} \implies \omega(t) = \frac{Fr}{I_z}t$$And then you can solve for the time at which rolling occurs. It's just a much more roundabout way ?:)
 
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  • #9
When time dependence is the question, it is more appropriate to use Newtion's equations than work and energy.

The in--plane motion of a rigid body constitutes a translation of the centre of mass, and a rotation about the centre of mass. The external force is kinetic friction, that acts at the point of contact of the ball with the ground. Its torque is ##\mu mg r##.

There are equations both for the translation and the rotation: For the translation:
##mdv/dt =-\mu mg## ----->##v=v_0-\mu g t##
For the rotation
##I d\omega /dt= \mu mgr ## ------> ##\omega=(\mu mgr/I) t ##
The moment of inertia of the ball about its centre of mass is ##2/5 mr^2##, so ##\omega = 5/2 \mu g t/r ##
When the ball starts to roll without slipping,
##v=\omega r##.
##v_0-\mu g t=5/2 \mu g t##.
Solve for t.
 

Related to Mechanics: A bowling ball is thrown and rolls with slipping

1. What is slipping in mechanics?

Slipping in mechanics refers to the motion of an object where the point of contact with the surface it is moving on is not stationary. In other words, the object is sliding or skidding rather than rolling smoothly.

2. How does slipping affect the motion of a bowling ball?

Slipping can significantly affect the motion of a bowling ball. It can cause the ball to lose speed and change direction, making it difficult to accurately predict its path. It can also cause the ball to spin instead of rolling, which can affect its trajectory and ultimately its final position.

3. What factors contribute to slipping in a bowling ball?

Several factors can contribute to slipping in a bowling ball, including the surface it is rolling on, the weight and material of the ball, and the force and angle at which it is thrown. Additionally, any imperfections or irregularities on the surface can also cause slipping.

4. How can slipping be prevented in bowling?

To prevent slipping in bowling, it is essential to have a smooth and even surface for the ball to roll on. This can be achieved by regularly cleaning and oiling the lanes. It is also crucial to choose the right ball weight and material for the specific lane conditions and to throw the ball with proper technique and force.

5. Can slipping be advantageous in bowling?

In some cases, slipping can be advantageous in bowling. For example, a skilled bowler may intentionally use slipping to create a spin on the ball, which can help it curve around obstacles and hit the pins more accurately. However, in general, it is best to avoid slipping for consistent and predictable results in bowling.

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