# Homework Help: Calculate the number of revolutions the disc makes

1. Dec 8, 2005

### kingyof2thejring

A disc of moment of inertia 29.1 kg m2 is made to rotate about an axis through its centre by a torque of T . The disc starts from rest and, after {t} s, has kinetic energy {k} J.

The torque is removed when the disc is rotating at 8.0 rad s-1 . An opposing torque of 13.2 N m is applied to slow the disc down.

Calculate the number of revolutions the disc makes after the second torque is applied, before it comes to rest.

how do i approch this problem (if numbers were involved) i would have to calculate the angular acceleration and then tention. after t seconds
any help would be apprecated thanks in advance

2. Dec 8, 2005

### Fermat

This is a problem in rotational kinematics.

Could you do it if it were a problem in linear kinematics, in other words, suppose you have a mass of 29.1 kg, with an initial speed of 8 m/s and a resisting force of 32.2 N, could you work out the distance travelled until it stopped ?

What (kinematic) eqns would you use to solve that problem ?

The eqns needed for the rotational problem are identical in form with their linear counterparts.

e.g.
s = ut + (1/2)at² - linear
θ = ωt + (1/2)αt² - rotational

If you know the eqns to use for the linear problem, then just convert them to their rotational equivelents.

3. Dec 8, 2005

### kingyof2thejring

hi, er... well... i've got the following data...

s=?
u=8
v=0
a=? (-8/t)
t=?

s= ut + 0.5at^2 ------> s= 4t

is the initial force equal to the resisting force. so that the force at the end equals 0

4. Dec 8, 2005

### Fermat

The resisting force is the decelerating force and is constant throughout the travel, s, of the mass.
Use newton's 2nd law, F = ma, to get a numerical value for the deceleration. call it a.

You need to use another kinematic eqn as well, though.

5. Dec 8, 2005

### Fermat

It's time I got some sleep (it's 1.00 a.m. here) so it's time I was off. Here's how to do it.

Use the eqn of motion,

v² = u² - 2as

where v is the final velocity (v = 0) and u is the initial velocity (u=8) and a is the deceleration (a = -2.2045 m/s²).

You will get a distance moved of,

s = 14.515 m
==========

Now all you have to do is write out the solution using rotational eqns of motion.

Substitute for,
s = θ
u = ω
a = α
The numerical values in your rotational eqns will be identical to those used in the linear eqns and the final angular displacement will be θ = 14.515 rads.
If you have been having any problems handling rotational eqns of motion, then after you've done it this way a few times, you will be able to just use the rotational eqns of motion straight away.