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).
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