# Sphere rolling down an incline problem.

1. Nov 5, 2014

### Brocoly

1. The problem statement, all variables and given/known data
A uniform solid sphere, of radius 0.20 m, rolls without slipping 6.0 m down a ramp that is inclined at 28° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?

2. Relevant equations
KE=1/2Iw^2
PE=mgh
I don't know what to use for this.

3. The attempt at a solution
Ma = MgSin(28)-Ff

2. Nov 5, 2014

### BvU

Hi veg, welcome to PF.
Your attempt at a solution has fallen off the map, and there is some more needed under 2) relevant equations.

I hope you've done a few slip/non slip slope exercises and are familiar with the combination of rolling and accelerating ?

In that case you know that he kinetic energy equation you listed applies to rotation about a stationary axis. Here, however, the axis is moving down the slope, which accounts for a kinetic energy due to linear motion too: KE = 1/2 m v2. So revise the list of eqns and show your attempt at solution , please :)

3. Nov 5, 2014

### Brocoly

1. The problem statement, all variables and given/known data
A uniform solid sphere, of radius 0.20 m, rolls without slipping 6.0 m down a ramp that is inclined at 28° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?

2. Relevant equations
w=(theta)/t
a=w/t
v=r(theta)/t
I don't know what to use for this.

3. The attempt at a solution
Acceleration = m(9.81m/s^2)(sin(28)) - Friction
At the end it did 6.0m/2pi(0.20m) revs over the time it took. (4.78rev/t)
from mgh=1/2mv^2+1/2Iw^2 i can get v^2=2gh/(1+I/mr^2)
I have no idea how to find the time from these variables.

Last edited: Nov 5, 2014
4. Nov 5, 2014

### BvU

Much better!

$\omega =\Delta \theta/t$ is for average angular velocity. Not good here: we have some constant acceleration and some constant angular acceleration, and the coupling is established by the non-slipping condition, which you kind of mention: $v = \omega r$.

Look up some equations for linear motion with constant acceleration plus some (surprisingly similar-looking) equations for angular motion with constant angular acceleration and see what you can use to get a grip on this nice exercise !

There might even be a shortcut possible: the time isn't asked for, so perhaps an energy balance (more equations still!) gets us to the answer in one fell swoop (provided it's a complete balance...) !

5. Nov 5, 2014

### Brocoly

Ok i think i got it.
v=rw
x=r(theta)
a=r(angular a)

(9.81)(sin(28))/(0.20)=angular a=23
(theta)=6/0.20=30
w^2=2(23)(30)
[w=37.1] is this correct?

6. Nov 5, 2014

### BvU

Acceleration = m(9.81m/s^2)(sin(28)) - Friction was better ! So $\alpha$ is alittle smaller than you think.

Friction is the one that causes the ball to rotate around its axis. Look up $\tau = I\alpha$ and $\tau = r \times F$.

At the risk of repeating myself:
Look up some equations for linear motion with constant acceleration plus some (surprisingly similar-looking) equations for angular motion with constant angular acceleration and see what you can use to get a grip on this nice exercise !

7. Nov 5, 2014

### Brocoly

Ok so what i did to try to find it is i took sin(28)*6 to find the height of at the start which is 2.82m
Then i did mgh=1/2mv^2 + 1/2Iw^2 which worked out to be 2gh=v^2
I got v as 7.438m/s and from that i did v/r=w
so i got w as 37.2m/s
I just need to convert it to rad/s

Last edited: Nov 5, 2014
8. Nov 6, 2014

### BvU

$mgh = {1\over 2} mv^2 + {1\over 2} I\omega^2\ \Rightarrow \ 2gh = v^2 + I\omega^2/m$ is good. That's the energy balance.

The $2gh=v^2$ that you worked out misses the rotation part and can't be right.
And then: re dimension: $v=\omega r$, so with v in m/s and r in m, the quotient is in radians/s, not m/s.

Since you have $v=\omega r$ you only need $I$ to crack this one. (No need to go via $v$ if they want $\omega$)