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## 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

**Tau**(torque) =

**F**(force)

**r**(radius) =

**T**(tension of string)

**r**(radius of wheel)

**α**(angular acceleration) =

**Tau**(torque)/

**I**(moment of inertia)

**a**(tangential acceleration) =

**r**(radius)

**α**(angular acceleration)

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

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