A mass is hung from a pulley. Suppose m = 241 g (including the mass holder) is r

Question

A mass is hung from a pulley.
Suppose m = 241 g (including the mass holder) is released from rest. The pulley has diameter d = 2.37 cm, and you measure the angular acceleration of the pulley a = 0.875 rad/s2. Assume
- The system is frictionless.
- The string does not slip on the pulley.
- The string is always perpendicular to the diameter of the pulley.

a) Calculate a, the linear acceleration of m as it falls.
m/s2

b) Calculate T, the tension in the string.
N

c) Calculate t, the torque exerted by the string on the pulley.
N-m

d) Calculate I, the moment of inertia of the system.
kg-m2

Explanation / Answer

The mass hung from pulley m = 0.241kg

the radius of the pulley R = 1.185cm or 0.01185m

the angular acceleration of the pulley = 0.875 rad/s^2

(a) The linear acceleration

     a = R = (0.01185)(0.875)

               = 0.010368 m/s^2

from Newton's law

    mg - T = ma

therefore tension

           T = mg - ma

               = (0.241)(9.8 - 0.010368)

               = 2.36 N

the torque actin gon the system is

    = TR

      = (2.36)(0.01185)

      =    0.0279 N.m

(d) The moment of inertia

          = I

therefore I = 0.0279 / 0.875

                = 0.032 kgm^2

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