# Homework Help: Angular Velocity of the wheel

1. Apr 18, 2014

### PC22

A uniform Steel wheel of diameter, D=3m and mass, 8500kg is supported by a low friction bearing, shown as the large black spot in the drawing. Around the outer surface of the wheel, a light wire rope is wound. From the end of this wire rope, a mass, M=1000kg is attached

Initially, both the wheel and the mass are at rest. The mass is released, and falls through a hight of 30m before hitting the earth's surface. As the mass falls, the tension in the wire rope turnes the wheel, i.e., the wire rope unswinds

2. Relevant equations
what is the speed of the mass just before it hits the ground ? and, what is the corresponding angular velocity of the wheel?

(assume the acceleration owing to gravity is equal to 9.81m/s^2)

3. The attempt at a solution

I've had a go at both questions but I am not sure if they are right, could some one please let me know if this is correct or where I have gone wrong ?

Third equation of motion
V=U+2as
initial velocity =U=0M/s
uniformed accelleration= 9.81M/s
distance traveled =t=30m

V=u+2as
V=0+2x9.81x30

V=588.6m/s

angular velocity

ω=v/r

ω=588.6/1.5

ω=392.4r/s

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2. Apr 18, 2014

### paisiello2

The acceleration is not g. It would be if the steel wheel had negligible mass compared to the mass M.

But imagine if the steel wheel were extremely large and massive and the mass M were very small. Would you think intuitively that the acceleration would still be g?

3. Apr 18, 2014

According to the principle of conservation of energy, the the speed of the block will be as shown in the picture.
The angular speed formula is correct, the problem is the speed.

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4. Apr 18, 2014

Hope it helped!

5. Apr 18, 2014

### PC22

AAAHH i'm an idiot I forgot to square root the velocity!!! so this is correct then??

Vroot2=u+2as
Vroot2=0+2x9.81x30
Vroot2=24.2m/s

angular velocity
ω=v/r
ω=24.2/1.5

ω=16.13r/m

6. Apr 18, 2014

Yes! Now it is correct!
Keep in mind that both velocities are squared in that particular equation

7. Apr 18, 2014

### paisiello2

Does the rotating wheel also have kinetic energy?

8. Apr 18, 2014

K=(1/2)Iw^2

I = moment of inertia= (1/2)mr^2
w = angular speed = v/r

K=1/2*1/2*m*r^2*(v/r)^2
K=1/4*m*v^2

When an object has translational as well as rotational motion, its total kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy.

9. Apr 18, 2014

### paisiello2

Yes, but you didn't include this in your estimate of v for mass M.

10. Apr 18, 2014

Ok let me try again.
The energy of the system must stay the same, therefore, the variation of energy of the block must be the same (symmetric) as the variation of the energy of the wheel.
The loss of potential energy of the block will result on the increase of the kinetic energy.

KE=PE
1/4mv^2=mgh
v^2=4gh
v=(4*9.81*30)^1/2
v=34,31m/s

11. Apr 18, 2014

w=v/r
w=34,31/1,5
w=22,87

12. Apr 18, 2014

### paisiello2

Now what happened to the translational kinetic energy? Your first attempt included it but excluded the rotational component of kinetic energy. Now for some reason you are including rotational but excluding translational.

13. Apr 18, 2014

But the wheel is attached isn't it? Therefore there would only be rotational kinetic energy..

14. Apr 18, 2014

### paisiello2

Attached to what?

You have to look at the energy of the entire system. Or if you only want to look at the wheel then you have to consider all forces that exert work on the wheel.

15. Apr 18, 2014

It says that the wheel is supported. There is no translational movement. Only rotational

16. Apr 18, 2014

### paisiello2

but the mass has translational velocity, yes?

17. Apr 18, 2014

But the wheel as a whole does not move through space. It only spins around it's center of mass.
Kinetic energy is only rotational.

18. Apr 18, 2014

### paisiello2

Yes, but it is attached to the falling mass which has a translational kinetic energy. you have to consider the system as a whole.

19. Apr 18, 2014

Yes, but when the block comes to rest, all of its translational kinetic energy is transformed into rotational kinetic energy on the wheel. That is what my previous calculations demonstrated

20. Apr 18, 2014

### paisiello2

but that is not what the question asked.

and even if it did you still calculated it wrong.