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A 3.00 kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform 8.00 kg cylindrical pulley. The inner diameter of the pulley is 60.0 cm, while the outer diameter is 1.00 m. The system is released from rest, and there is no friction at the axle of the pulley. Find (a) the acceleration of the stone, (b) the tension in the wire, and (c) the angular acceleration of the pulley.

for part a i did
I of the pulley is
I = m*(r1^2 + r2^2)/2)
I = 8*(0.5^2 + 0.3^2)/2)
I =1.36 kgm^2

a = 3*9.81/(3 + 1.36/0.5^2)
a = 3.48 m/s^2 <--- acceleration of the stone

part B
á = a/R = 3.48/0.5 = 6.96 rad/s^2 <--- angular acceleration

Part C
T = mg - ma = 3(9.81-3.48)=18.99 N <--- tension

B and C is switched anwsers

• Oh my - sign ! -

The inner radius I assume means that the axle does not spin. Therefore your moment of inertia has a negative, not positive sign
I = 8*(0.5^2 - 0.3^2)/2)
I =.64 kgm^2

stone F = m a
T up
mg down
a = (mg-T)/ m = (3*9.81 -T)/3

wheel Torque = I alpha
alpha = .5 T/.64
a = alpha * r
a = .25 T/.64

so
(3*9.81 -T)/3 = .25 T/.64

solve for T
go back and solve for a
and for alpha = a/r

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