A cord of negligible mass is wrapped around a frictionless pulley of mass M and

Question

A cord of negligible mass is wrapped around a frictionless pulley of mass M and radius R. The pulley rotates about a fixed axle that passes through its center. A bucket of mass m hangs from the cord. Write force and torque equations for the objects Bucket Falling (Force equation) Pulley Rotating (Torque equation) The mass of the pulley is M = 5 kg, its Moment of Inertia = 0.1 kgm, the mass of the bucket is m = 1 kg, and the radius of the pulley is 0.2 m. Combine the two equations above to show that the magnitude of the linear acceleration of the bucket is 2.8 m/s2. Determine the angular acceleration of the pulley. Determine the tension T in the cord

Explanation / Answer

let T is the tension in the string.

Force equation,

net force acting on bucket, Fnet = m*g - T

m*a = m*g - T

T = m*g - m*a -----(1)

pulley roatating,

net torque acting on the pulley = T*R

I*alfa = T*R   ----(2)

on combining equations 1 and 2

I*alfa = (m*g - m*a)*R

I*a/R = (m*g - m*a)*R

I*a/R^2 = m*g - ma*

m*a + I*a/R^2 = m*g

a*( m + I/R^2) = m*g

a = m*g/( m + I/R^2)

= 1*9.8/(1 + 0.1/0.2^2)

= 2.8 m/s^2 <<<<<<<<<<-----------------------Answer

angular aceleration, alfa = a/R

= 2.8/0.2

= 14 rad/s^2 <<<<<<<<<<-----------------------Answer

T = m*g - m*a

= 1*9.8 - 1*2.8

= 7.0 N <<<<<<<<<<-----------------------Answer

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