A string, wrapped around a thin walled hollow wheel (as shown) with a radius of

Question

A string, wrapped around a thin walled hollow wheel (as shown) with
a radius of 0.25 m is pulled with a constant force F. The moment of inertia
of the wheel about its axle is 0.050 kg-m2. The wheel starts from rest and
reaches a rotational rate of 16 rad/s in 4 seconds. [NOTE: The moment of
inertia of a thin walled hollow cylinder of mass M and radius R about its
central axis is given by I = MR

A string, wrapped around a thin walled hollow wheel (as shown) with a radius of 0.25 m is pulled with a constant force F. The moment of inertia of the wheel about its axle is 0.050 kg-m2. The wheel starts from rest and reaches a rotational rate of 16 rad/s in 4 seconds. [NOTE: The moment of inertia of a thin walled hollow cylinder of mass M and radius R about its central axis is given by I = MR 2] Find: the angular acceleration of the wheel the magnitude of the force F. the torque produced by F the angular displacement of the wheel as it went from rest to a rate of 16 rad/s. the mass of the wheel the work performed on the wheel as it accelerated to 16 rad/s from rest show the work performed on the wheel is equal to the change of its kinetic energy

Explanation / Answer

r = 0.25 m

I = 0.05 kg.m^2

wo = 0 at t = 0

w = 16 rad/s at t = 4s

a) alfa = (w - wo)/t = 16/4 = 4 rad/s^2

b) Torque, T = I*alfa

F*r = I*alfa

F = I*alfa/r

= 0.05*4/0.25

= 0.8 N

c) theat = 0.5*alfa*t^2

= 0.5*4*4^2

= 32 radians

d) T = r*F = 0.25*0.8 = 0.2 N.m

e) I = M*r^2

M = I/r^2

= 0.05/0.25^2

= 0.8 kg

f) work done = T*theta

= 0.2*32
= 6.4 J

g)

change in kinetic energy = 0.5*I*(w^2-wo^2)

= 0.5*I*w^2

= 0.5*0.05*16^2

= 6.4 J

workdone = change in kinetic enrgy

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