?3. (5 pts) A can whose rmoment of inertia (I) is 0.2 kg-m2 is rolling along a h

Question

?3. (5 pts) A can whose rmoment of inertia (I) is 0.2 kg-m2 is rolling along a horizontal floor at m/'S. This can has a mass of 1.6 kg and a radius of 50 cm. (a) What is the translational kinetic energy of the can? (b) What is the angular velocity (o) of the can? (c) What is the rotational kinetic energy of the can? (d) What is the total kinetic energy of the can? (e) If the can runs into a bumper, what work must the bumper do to stop it? (f) What are the linear and angular momenta of the can initially?

Explanation / Answer

Given

mass of can m = 1.6 kg , rolling along horizontal with velocity v = 10 m/s

moment of inertia I = 0.2 kg-m^2, radius r = 0.5 m

a) translational kinetic energy k.e1 = 0.5*m*v^2

k.e1 = 0.5*1.6*10^2 = 80 J

b) angular velocity W = ?

we have relation V = r*W

W = V/r = 10/0.5 = 20 rad/s

c) rotational kinetic energy is  

k.e2 = 0.5*I*W^2

k.e2 = 0.5*0.2*20^2 J

k.e2 = 40 J

d) Total k.e of the can  

k.e = k.e1+k.e2

k.e= 80+40 = 120 J

e) work done by the bumper so that the can comes to rest is  

from work energy theorem the work done = change in k.e

Work doen = (120-0)= 120 J

f)

linear momentum P = m*v = 1.6*10 kg m/s = 16 kg m/s

angular momentum L = I*W = 0.2*20 = 4 kg m/s2

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