# Velocity of mass connected to a fixed rotating wheel with a massless string

## Homework Statement

A mass of 0.8kg is attached to a massless string that passes over a pulley and is wrapped around a wheel of radius 0.5m, and rotational inertia 1.0 kgm^2. The wheel can rotate freely. The mass is initially held stationary at a height of 1.5m above the ground. When the mass is released it falls to the floor, causing the wheel to rotate via the attached string. What is the speed of the mass when it reaches the ground?

(Just to make it clearer, as a diagram is included; the wheel is fixed on a block, rotating horizontally, with the string going through a pulley, then turned 90 degrees (vertical to the ground). This string is then attached to the mass, where the mass is initially 1.5m off the ground). ## Homework Equations

α(angular acceleration) = Tau(torque)/I(moment of inertia)
I = Mr^2

Vf^2 = Vi^2 + 2ah

## The Attempt at a Solution

TR = Iα
= M(g - a)= Iα (since total acceleration of the string is in the opposite direction to tension)
=0.5(0.8*9.8 - 0.8*a) = α (because I is equal to 1 kgm^2)
(0.5*r)*(7.84 - 0.8*a) = a
0.5*(7.84 - 0.8*a) = 4a
7.84 - 0.8a = 8a
8.8a = 7.84
a= 0.89ms^-2 --> I didn't feel confident with this, so I didn't coninue on with it. I tried working backwards from the speed given, knowing that the inital speed is zero, and that the height is 1,5m, but there must be some concept that I don;t understand.

Other times I tried this, I found a circular result, or that I would get acceleration equal to 9.8ms^-2. I'm not sure if I should somehow use the mass of the wheel, calculating it using moment of inertia and the radius of wheel. The correct solution that the past exam paper includes is speed = 0.5ms^-1

Last edited:

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Doc Al
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(Just to make it clearer, as a diagram is included; the wheel is fixed on a block, rotating horizontally, with the string going through a pulley, then turned 90 degrees (vertical to the ground). This string is then attached to the mass, where the mass is initially 1.5m off the ground).
Your diagram shows the string connected to something on top of the block. I assume that's just an error.

TR = Iα
= M(g - a)= Iα (since total acceleration of the string is in the opposite direction to tension)
=0.5(0.8*9.8 - 0.8*a) = α (because I is equal to 1 kgm^2)
(0.5*r)*(7.84 - 0.8*a) = a
OK.
0.5*(7.84 - 0.8*a) = 4a
Redo this step.

You can also use conservation of energy to solve this problem.

The thing on the top of the block is the wheel from which the string is wrapped around. Sorry if I wasn't clear enough. :)

Doc Al
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The thing on the top of the block is the wheel from which the string is wrapped around. Sorry if I wasn't clear enough. :)
I see a string that attaches to the hanging mass, passes over a pulley, then attaches to some unknown thing. But I assume from your description (and solution) that the string just wraps around the pulley and doesn't attach to some other thing. Right?

Edit: Never mind! Now I see what you are saying. That "unknown thing" is the wheel; the pulley is presumed massless. :uhh:

But my comments in the previous post remain.

Reviewing the step 0.5*(7.84 - 0.8*a) = 4a, I found where i made a mistake. I now know that that should be (7.84 - 0.8*a) = 4a
Which simplifies to:

7.84 = 4.8a
a= 1.63 ms^-1
vi= 0ms^-1
h= 1.5m
Vf^2 = Vi^2 +2ah
Vf^2 = 0 + 2*1.63*1.5
Vf^2 = 4.9 m^2*s^-4
Vf= 2.21 ms^-1 ---> which is unfortunately wrong according to the answer. :(

Doc Al
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I've also tried using the conservation of energy as you suggest.

Ki + Ui = Kf + Uf
0 + mgh= 0.5mv^2 + 0.5Iω^2 + 0
0.8*9.8*1.5 = 0.5*0.8*v^2 +0.5ω^2
11.76r^2 = r^2*0.4V^2 + 0.5(ωr)^2
2.94= 0.1v^2 + 0.5v^2
v^2 = 4.9 m^2*s^-2
v = 2.21 ms^-1
I guess perhaps the answer given is wrong, or maybe a crazy coincidence.

Doc Al
Mentor
Does it give an answer? Or is are you using an online system?

It's a question from a past exam paper from my uni, and they supplied the answers (without worked solutions) online.