# Angular Acceleration and Torque

• jakec
In summary, the stuntman grabs a rope connected to a 400kg cylinder and falls from a balcony. The torque applied by the stuntman causes the stuntman and the cylinder to accelerate towards the ground.
jakec

## Homework Statement

To lower himself from a balcony an 80kg stuntman grabs a rope connected to a 400kg cylinder with a 1.2m diameter that is free to rotate about its axis of symmetry. What is the stuntman's acceleration as he falls?

## Homework Equations

I missed this on a homework assignment. I know the correct answer is 2.86m/s2 but I can't find what I did wrong.

## The Attempt at a Solution

moment of inertia for a cylinder: I=1/2 Mr2

torque applied by stuntman = mgr = 80kg * 9.8m/s2 * 0.6m = 470.4Nm
Using t = Iα
470.4Nm = (0.5 * 400kg * 0.6m2
α = 470.4Nm / (0.5 * 400kg * 0.6m2) = 6.53rad/s2
a = αr = 6.53rad/s2 * 0.6m = 3.92m/s2

What am I missing?

jakec said:
torque applied by stuntman = mgr = 80kg * 9.8m/s2 * 0.6m = 470.4Nm
Hello. This is not the correct expression for the torque. The tension in the rope is not equal to the weight of the man. The tension is one of the unknowns. That means you'll need an additional equation besides ##\tau = I \alpha##. Be sure to draw free body diagrams for both the cylinder and the man.

The force on the stuntman is accelerating the stuntman as well as turning the cylinder. You need to work out how to partition the force into those two parts.

OK, so this is what I have now. I'm still a little off from the answer I should get. Any suggestions of what I can do better? (hopefully in time for finals tomorrow)

T= tension
ζ = torque
W = weight of stuntman

∑ζ = Iα
Since the tension in the cord is the only force providing torque:
Tr = Iα
Tr = (1/2 Mr2)(a/r)
T = 1/2 Ma

ΣF = ma
W - T = ma
Substituting from above:
W - (1/2 Ma) = ma
a = (2mg) / (2m +M)

a = (2 * 80kg * 9.8m/s2) / (2 * 80kg + 400kg) = 2.8 m/s2

Your work looks correct to me. I also get 2.8 m/s2.

Great! Thanks for the help!

## 1. What is angular acceleration?

Angular acceleration is a measure of the rate at which the angular velocity of an object changes over time. It is typically measured in radians per second squared.

## 2. How is angular acceleration related to linear acceleration?

Angular acceleration is related to linear acceleration through the radius of rotation. The formula for converting between the two is a = αr, where a is linear acceleration, α is angular acceleration, and r is the radius of rotation.

## 3. What factors affect angular acceleration?

The main factors that affect angular acceleration are the net torque applied to an object and the moment of inertia, which is a measure of an object's resistance to rotational motion. Other factors that may affect angular acceleration include the shape and distribution of mass of an object.

## 4. How is torque related to angular acceleration?

Torque is directly proportional to angular acceleration. The greater the torque applied to an object, the greater the angular acceleration will be. This relationship is described by the formula τ = Iα, where τ is torque, I is moment of inertia, and α is angular acceleration.

## 5. What is the difference between angular acceleration and angular velocity?

Angular acceleration is a measure of how quickly the angular velocity of an object changes over time, while angular velocity is a measure of how fast an object is rotating at a given moment. In other words, angular velocity is the rate of change of angular displacement, while angular acceleration is the rate of change of angular velocity.

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