# Simple Angular Velocity Problem

• Dorothy Weglend
In summary, the problem involves a dog running around a stationary turntable with angular velocity w = 0.750 rad/s. The dog sees a bone 1/3 revolution away and the turntable then begins to turn with angular velocity w = 0.015 rad/s. The questions ask how long it will take for the dog to reach the bone, as well as how long it will take for the dog to reach the bone again if he overshoots it. The answer involves using the frame of the rotating disc, and taking the point where the dog is when the turntable starts as the reference for the angular measurement. The relative speed of the dog is the speed of the dog running on the disc plus the rotation of the
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

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

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.

Last edited:
ponjavic said:
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.

andrevdh said:
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

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.

## 1. What is angular velocity?

Angular velocity is a measure of how fast an object is rotating around a fixed point. It is typically measured in radians per second (rad/s) or degrees per second (deg/s).

## 2. How is angular velocity different from linear velocity?

Angular velocity is a measure of rotational speed, while linear velocity is a measure of straight-line speed. Angular velocity takes into account the distance from the rotation axis, while linear velocity does not.

## 3. How do you calculate angular velocity?

Angular velocity can be calculated by dividing the change in angle (in radians or degrees) by the change in time. The formula is: angular velocity = change in angle/change in time.

## 4. What is the relationship between angular velocity and angular acceleration?

Angular acceleration is the rate of change of angular velocity. In other words, it measures how quickly the angular velocity is changing. The formula for angular acceleration is: angular acceleration = change in angular velocity/change in time.

## 5. How can I use angular velocity in real life?

Angular velocity is used in various real-life applications, such as in the design of car engines, amusement park rides, and sports equipment like bicycles and gyroscope toys. It is also important in understanding the motion of celestial bodies and in studying the Earth's rotation.

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