# Simple concept help: Pulley and Net Torque

## Homework Statement

A ball (hollow sphere) of radius .11m and mass .444kg is mounted on a frictionless axle. A massless cord is wrapped around the ball supporting an object of mass .02kg. Find the angular acceleration of the wheel, linear acceleration of the object and tension in the cord.

## The Attempt at a Solution

$$\Sigma$$F=Fg+T=m*a
-.2+T=.02a

$$\Sigma$$$$\tau$$=F*r=I*$$\alpha$$
-T*.11=(.0036*a)/.11
T = -.2975*a

-.2 + .2975*a = .02*a
-.2 = -.2775a
.72 = a

That's what I tried, but it's wrong.
Can anyone help?

## The Attempt at a Solution

#### Attachments

• net torque.png
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PhanthomJay
Homework Helper
Gold Member

## Homework Statement

A ball (hollow sphere) of radius .11m and mass .444kg is mounted on a frictionless axle. A massless cord is wrapped around the ball supporting an object of mass .02kg. Find the angular acceleration of the wheel, linear acceleration of the object and tension in the cord.

## The Attempt at a Solution

$$\Sigma$$F=Fg+T=m*a
-.2+T=.02a
0.2 -T = .02a
$$\Sigma$$$$\tau$$=F*r=I*$$\alpha$$
-T*.11=(.0036*a)/.11
T*.11 = .0036a/.11
T = -.2975*a
T = .2975a
-.2 + .2975*a = .02*a
-.2 = -.2775a
.72 = a

That's what I tried, but it's wrong.
Can anyone help?

## The Attempt at a Solution

You must watch your signage. The acceleration is downwards. Choose the downward and clockwise directions as positive, for compatability and simplicity.