A thin uniform rod of length d = 1.56 m and mass M = 0.88 kg is lying on a frict

Q: A thin uniform rod of length d = 1.56 m and mass M = 0.88 kg is lying on a frict

Given M = 0.88 kg d = 1.56 m F = 1500 N dt = 1 ms = 1*10^-3 s a) we know, net torque = rate of change of angular momentum T = dL/dt T*dt = dL T*dt = I*(w - wo) T*dt = I*w (since, wo = 0) w = T*dt/I = F*(d/2)*dt/(M*d^2) = F*dt/(2*M*d) = 1500*1*10^-3/(2*0.88*1.5) = 0.568 rad/s

ID: 1471289 • Letter: A

Question

A thin uniform rod of length d = 1.56 m and mass M = 0.88 kg is lying on a frictionless horizontal surface. One end of the rod is given a sudden impulsive force of magnitude F = 1500 N in a direction perpendicular to the length of the rod (see the figure below). The sudden impulsive force acts for a very short interval of time t = 1 ms. After the application of the impulsive force the rod starts to move (translational motion) and rotate.

Because the time interval for the applied impulsive force is short, you can assume that the rod moves/rotates through a negligible distance/angle during the application of the force (i.e. during the time interval t). Also, note that the figure is an overhead-view of the rod on the horizontal surface.

(a) What is the angular speed of the rod after the sudden impulsive force is applied?
=  rad/s

(b) What is the translational speed of the center of mass of the rod after the sudden impulsive force is applied?
vCM =  m/s

(c) After the rod has completed two full revolutions, what is the position of the center of mass of the rod on the xy-coordinate system shown in the figure?  You can assume that the CM of the rod was initially at the origin of this coordinate system, and that the positive x and y-directions are to the right and up, respectively.

Explanation / Answer

Given

M = 0.88 kg

d = 1.56 m

F = 1500 N

dt = 1 ms = 1*10^-3 s

a) we know, net torque = rate of change of angular momentum

T = dL/dt

T*dt = dL

T*dt = I*(w - wo)

T*dt = I*w (since, wo = 0)

w = T*dt/I

= F*(d/2)*dt/(M*d^2)

= F*dt/(2*M*d)

= 1500*1*10^-3/(2*0.88*1.5)

= 0.568 rad/s <<<<<<<<<<<<<<<<<<<<<<<------------Answer

b) Apply Impulse = change in momentum

F*dt = M*(v - vo)

F*dt = M*vcm (here, vo = 0)

vcm = F*dt/M

= 1500*10^-3/0.88

= 1.7 m/s <<<<<<<<<<<<<<<<<<<<<<<------------Answer

c) time taken for two revolutions, t = 2*T

= 2*2*pi/w

= 2*2*pi/0.568

= 22.1 s

horizontal poistion, x = vcm*t

= 1.7*22.1

= 37.6 m <<<<<<<<<<<<<<<<<<<<<<<------------Answer

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A thin uniform rod of length d = 1.56 m and mass M = 0.88 kg is lying on a frict

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