A uniform circular disc of mass 3.11 kg and radius 0.274 m rests on a horizontal

Question

A uniform circular disc of mass 3.11 kg and radius 0.274 m rests on a horizontal frictionless surface; i.e., the disk is allowed to slip on the surface. A constant force of 7.25 N is applied to the end of a string that is wrapped around the disk causing the disk to rotate about a vertical axis and to translate in the horizontal direction. Find the linear acceleration of the disk. Answer in units of m/s^2 . What is the total energy of the system 20 s after the force is applied? Answer in units of J.

Explanation / Answer

work done by Torque

T=f*d

=7.25*0.274

T=1.98Nm

inertia of disc, this is 1/2mr^2 = 0.5 x 3.11 x 0.274^2

inertia of disc= 0.116 kg-m^2

Torque = inertia x alpha (rad/s^2)

1.98Nm/0.116 kg-m^2 =alpha(rad/s^2)

17.06rad/sec^2=alpha(rad/s^2)

w=17.06 x 20=341.2rad/sec

w=341.2rad/sec

Linear accleration of the disc=T/mass of string

=7.25 N /3.11

a)Linear accleration of the disc=2.331 m/s^2

b)Total energy = 1/2mv^2 + 1/2Iw^2

=1/2(3.1*2.331^2+0.116*341.2^2)

=0.5(16.844+13504)=0.5 (13521.26)

Total Energy=6760.63J

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