A wheel rotates about a horizontal axis, as shown in the figure. It has an outer

Question

A wheel rotates about a horizontal axis, as shown in the figure. It has an outer radius R1 = 21.2 cm and mass M1 = 4.74 kg. Attached to it concentrically is a hub (shown cross-hatched) of radius R2 = 4.91 cm and mass M2 = 1.11 kg. The axle has negligible radius and mass. Both the wheel and the hub are solid and have uniform density. Suspended from a massless string that is wound around the hub is a weight of mass m = 472 g.

a) What is the acceleration of the hanging weight after it is released?

I got this to be .1024 m/s^2 which is right.

b) When the weight has fallen a distance 61.0 cm, what is the kinetic energy of the rotating wheel?

I can't get this part though.

Explanation / Answer

b)

M1 = 4.74 kg, R1 = 0.212 m
M2 = 1.11 kg, R2 = 0.0491 m

I1 = moment of inertia of wheel = M1*R1^2/2 = 0.1065 Kg.m^2

I2 = moment of inertia of hub = mr^2/2 = 0.00134 Kg.m^2

I = I1 + I2 = 0.10784 Kg.m^2

a = 0.1024 m/s^2

let t is time taken to fall down

s = u*t + 0.5*a*t^2

0.61 = 0 + 0.5*0.1024*t^2

t = 2*0.61/0.1024 = 11.914 s


Torque on the rotating object, T = R1*m*a = 0.212*0.472*0.1024


Torque, T = 0.010246 N.m

i think u have found moment of inertis of the system in part a.

Torque, T = I*alfa

alfa = T/I = 0.010246/0.10784 = 0.095 rad/s^2

let w1 is the initial angular velocity and w2 is the final angular velocity

w2 = w1 + alfa*t

w2 = 0 + alfa*11.914

w2 = 0.095*11.914 = 1.132 rad/s


kinetic energy of the system after 11.914 s

KE = 0.5*I*w2^2 = 0.5*0.10784*1.132^2 = 0.069 J

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