# Homework Help: Currently studying for finals, need help with torque

1. Dec 10, 2012

### penguinnnnnx5

1. The problem statement, all variables and given/known data

A ball of mass m and radius r (Icm = 2mr^2 /5) is released from rest, and rolls without slipping down a ramp.
(a) Draw a free body force diagram for the ball. [Hint: careful—there has to be friction. (Why?) Further, it is static friction, and not kinetic friction. (Why?)]
(b) Use ΣF = macm and Στcm = Icmαcm to find the (linear) acceleration of the ball down the ramp. [Ans: a = (5gsinθ)/7]
(c) Use your result from part (b) to find the linear speed of the ball after it rolls a distance L down the ramp. Check your answer by using conservation of energy.
(d) Solve for the friction force acting on the ball. [Ans: f = (2mgsinθ)/7.]
(e) Solve for the normal force, N.

The entire problem can be found here, diagram included

2. Relevant equations
Στcm = Icmcm = Fr
ΣF = m*acm
KE = (1/2)Icm*ω + (1/2)mv^2
PE = mgh

3. The attempt at a solution

a. Pretty straight forward, Force of friction opposes force of gravity in x direction and both are equal to each other. Static friction allows the ball to roll because when the ball does not slip, the portion touching the ramp is actually stationary (I can't think of a better word). If it were kinetic friction, then it is implied that the ball is sliding down the ramp.

b. This is where I have a problem. My past notes show the "correct" steps, but there is one step at the very beginning that I do not understand.

Στ = Iω + mr^2 *ω <-- Why is there that mr^2ω? If it's related to kinetic energy, then what happened the (1/2)s?
= (2/5)mr2ω + mr2ω = (7/5)mr2ω = (7/5)mrω

ΣF = ma = Fg,x - Ff = 0 ∴ Ff = Fg,x = mgsinθ

mgsinθ *r = (7/5)mra
a = (5/7)gsinθ

c. Did I do this correctly?
KE = (1/2)(2/5)mv2 + (1/2)mv2 = (7/10)mv2
PE = mgh = mgLsinθ
mgLsinθ = (7/10)mv2
v2 = (10/7)gLsinθ

d.
rFf = Iα = (2/5)mra
Ff = (2/5)ma = (2/5)(5/7)mgsinθ = (2/7)mgsinθ

e.
Ff = Nus = mgcosθ *us
N = mgcosθ

Anyway, what I really need help on is the first step of b. As I said, we already went over this problem in my discussion class, but I just don't understand the first step of b. Thank you for your time!

2. Dec 10, 2012

### TSny

Note that the above equation is not correct dimensionally. Should $\omega$ be replaced by $\alpha$? It looks like this is an attempt to set up torques about the point of contact of the sphere with the surface and the extra term with the $mr^2$ is due to the parallel axis theorem in shifting from the cm to the point of contact. But the question asks you to set up the equation using the cm as axis for the torques:

$\sum\tau_{cm} = I_{cm}\alpha$.
$a$ is not zero. You want to find $a$.

Last edited: Dec 10, 2012
3. Dec 10, 2012

### penguinnnnnx5

Ahhh, that makes more sense. I completely forgot about the parallel axis theorem. Thank you! It all makes sense now.