A 4.00kg stone is tied to a thin, light wire wrapped around the outer edge of th

Question

A 4.00kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform 10.0kg cylindrical pulley shown in the figure below (Figure 1) . The inner diameter of the pulley is 50.0cm , while the outer diameter is 1.00m . The system is released from rest, and there is no friction at the axle of the pulley.

Find the acceleration of the stone.

Part B

Find the tension in the wire.

Find the angular acceleration of the pulley.

a =   m/s2   A 4.00kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform 10.0kg cylindrical pulley shown in the figure below (Figure 1) . The inner diameter of the pulley is 50.0cm , while the outer diameter is 1.00m . The system is released from rest, and there is no friction at the axle of the pulley. a = m/s2 Part B Find the tension in the wire. T = N Find the angular acceleration of the pulley. ? = rad/s2

Explanation / Answer

Torque = R*T = I*? = I*a/R

where R is the outer radius of the pulley
T is the tension off the wire
I = moment of inertia of the pulley
? = angular acceleration
a = tangential acceleration

the tension T is mass of the stone*( g - a).
this makes the equation to
(mg - ma)*R = I*a/R ( m = mass of the stone)
mg - ma = Ia/R^2 --> solve for a
mg = a(m + I/R^2)
a = mg/ (m + I/R^2)

I of the pulley is
I = m*(r1^2 + r2^2)/2)
I = 10*(0.5^2 + 0.25^2)/2)
I = 1.56 kgm^2

T-mg=-ma

T=mg-ma
1.56a = 4g - 4a
a= 4 g / 5.56 = 7.05m/s/s
a = 7.05 m/s^2 <--- acceleration of the stone

? = a/R = 7.05/0.5 = 14.11 rad/s^2 <--- angular acceleration

T = mg - ma = 4(9.81 - 7.05) = 11.04 N <--- tension

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