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...
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...
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...