Simple Angular Velocity Problem

[Dorothy Weglend]

A new topic for me, angular velocity. I haven't been able to solve this problem. Perhaps someone can help:

A dog is running around a stationary turntable, with angular velocity w = 0.750 rad/s. The dog sees a bone 1/3 rev away. At that instant, the turntable begins to turn with angular velocity w = 0.015 rad/s.

a) How long will the dog take to reach the bone?
b) If the dog overshoots the bone, and continues around the moving turntable, how long before he reaches the bone once again.

Measure all times from the point at which the turntable starts to move.

I thought I would solve this in the frame of the turntable, which makes it quite simple, but I get different answers from those supplied on the my sheet.

Speed of dog in frame of turntable:

w = 0.750 rad/s - 0.0150 rad/s = 0.735 rad/s

Then I use theta = wt.

Bone is 1/3 revolution away:

t = theta/w = (2 pi/3)/0.735 = 2.85 s

Time once around the turntable:

t = (2 pi)/0.735 = 8.55 s

Add these two to get total time from start back to bone: 11.4 s

Answers on the handout are 2.88 s for the time to the bone, and 12.8 seconds for the time to come back around once again.

Seems a little large for rounding errors.

I thought ignoring the acceleration might be the problem, but there isn't enough information to calculate that.

Well, thanks for any pointers.

Dorothy

 

[ponjavic]

You are absolutely right in trying to solve the problem using the frame of the rotating disc, however. What is the speed of the dog relative to the disc?

It clearly states that the dog is running around the table, so his actual speed is that of the running it does on the disc + the rotation of the disc.

Thus its relative speed is?

Also there is no accelleration at is it instantaneously moving at the specified angular velocity

 

[andrevdh]

Taking the point where the dog is when the turntable starts as the reference for the angular measurement. For the dog

\theta _d = \omega _d t

for the bone

\theta _b = \omega _b t + \theta _o

equating brings one to

t_1 = \frac{\theta _o}{\omega _d - \omega _b}

for the second meeting

t_2 = \frac{\theta _o + 2\pi}{\omega _d - \omega _b}

so I also doubt the answers on the handout.

 

[Dorothy Weglend]

You are absolutely right in trying to solve the problem using the frame of the rotating disc, however. What is the speed of the dog relative to the disc?

It clearly states that the dog is running around the table, so his actual speed is that of the running it does on the disc + the rotation of the disc.

Thus its relative speed is?

Also there is no accelleration at is it instantaneously moving at the specified angular velocity

Thanks, Ponjavic. I could have been clearer. There is a diagram that shows the dog running around the turntable, i.e., on the ground, outside of the turntable, not on it.

 

[Dorothy Weglend]

Taking the point where the dog is when the turntable starts as the reference for the angular measurement. For the dog

\theta _d = \omega _d t

for the bone

\theta _b = \omega _b t + \theta _o

equating brings one to

t_1 = \frac{\theta _o}{\omega _d - \omega _b}

for the second meeting

t_2 = \frac{\theta _o + 2\pi}{\omega _d - \omega _b}

so I also doubt the answers on the handout.

Thanks andrevdh, I appreciate it. Thanks also for including your solution. That was very instructive for me!

Dorothy

 

[andrevdh]

Well, it just confirms your simpler approach.

After they dog caught up with the bone for the first time he will catch up with it for the second time when:

\theta _d - \theta _b = 2\pi

this is especially true since the dog starts out \frac{2\pi}{3} before the bone and his angular distance from the bone will decrease due to the fact that he has a larger angular velocity than the bone. Therefore we have that

\omega _d t_2 - \omega _b t_2 - \theta _o = 2\pi

which give the last equation.