Question
Everyday application: Suppose a yo-yo has a center shaft that has a 0.250 cm radius and that its string is being pulled.
  1. If the string is stationary and the yo-yo accelerates away from it at a rate of 1.50 m/s21.50 \textrm{ m/s}^2, what is the angular acceleration of the yo-yo?
  2. What is the angular velocity after 0.750 s if it starts from rest?
  3. The outside radius of the yo-yo is 3.50 cm. What is the tangential acceleration of a point on its edge?
Question by OpenStax is licensed under CC BY 4.0
Final Answer

a) 600 rad/s2600 \textrm{ rad/s}^2

b) 450 rad/s450 \textrm{ rad/s}

c) 21.0 m/s221.0 \textrm{ m/s}^2

Solution video

OpenStax College Physics, Chapter 10, Problem 9 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We have the inner shaft of the yo-yo here and the string is wrapped around it. It has some radius of 0.25 centimeters which we convert into meters by multiplying by ten to the minus two. There is an outer radius that we'll use for part C to calculate the tangential acceleration of a point on the outside of the yo-yo. The tangential acceleration we're told is 1.5 meters per second squared. That is a point that's right here on the shaft edge. The outside diameter is three and a half centimeters. Now for part A we have to figure out the angular acceleration given this tangential acceleration. So we take 1.5 meters per second squared divided by the radius of the shaft at 0.25 times ten to minus two meters and we get 600 radians per second squared. For part B, we want to know what will the angular velocity of the yo-yo be after it has accelerated for 0.75 seconds. So we take the initial angular velocity of zero and add to that the angular acceleration of 600 radians per second squared, multiplied by three quarters of a second. This gives us 450 radians per second. Then the tangential acceleration of a point on the outside of the yo-yo is going to be its angular acceleration multiplied by the outer radius. So it's 600 radians per second squared times three and a half centimeters and this gives 21.0 meters per second squared.

Comments

for part C, why do you multiply the outside radius by the angular acceleration solved for the inside radius?

Hi Jairus, thanks for the good question. Imagine a line from the center to the outside radius. It's easier to picture angular displacement, so let's talk about that. Angular displacement is the amount of angle that a radius moves as the disk rotates. Compare the angle moved by a line from the center to half-way along the radius to the angle moved by a line that is the entire radius -> they're the same! Points along a radius have the same angular displacement, and they also likewise have the same angular velocity and angular acceleration. This is why the angular acceleration for the inside radius is the same as the angular acceleration of the outside radius, so we can multiply the outside radius by the angular acceleration of the inside radius (which is the same angular acceleration as for the outside).
Hope this helps,
Shaun