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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

**f**orce of

**f**riction ##F_f=mg\mu##, where ##m## is the ball's

**m**ass. 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

**a**cceleration. 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

**k**inetic

**e**nergy of the ball, ##E_{k_1}## is the

**k**inetic

**e**nergy of the ball at the moment, it stop slipping and ##E_{r_1}## is its

**r**otational

**e**nergy 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

**i**nertial 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

**w**ork of the force of

**f**riction (##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 (

**s**patium 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.