A solid disk (mass = 5 kg, R=0.7 m) is rolling across a table with a translation

Question

A solid disk (mass = 5 kg, R=0.7 m) is rolling across a table with a translational speed of 10 m/s.
Note: Be careful to avoid roundoff errors in this problem.

a.) What must the angular speed of the disk be?
14.28  rad/s
b.) What is the rotational kinetic energy of the disk?
124.9 J
c.) What is the total kinetic energy of the disk?
_____ J
d.) The disk then rolls up a hill of height 2 m, where the ground again levels out. Find the translational and rotational speeds now.
_____  m/s
_____  rad/s

need help with c and d. Thanks

Explanation / Answer

The total kinetic energy of disc = rotational energy + translational energy =1/2 x I x w^2 + 1/2 x Mx v^2

and I = MxR^2/2 ( this can be dervied from simple integration)

I = 5 x .7 ^2 /2 =1.225 and w = V / R =10/.7 = 14.28 m/sec

Rotatinal energy =1/2 x 1.225 x 14.28 ^2 = 125

translational kinetic energy = 1/2 x5 x 10^2 = 250

total kinetic energy = 375 J Answer C

When disc roll from 2 m height

then total kinetic energy = previous kinetc energy - mgh ( potential energy)

=375- 98 =277

277 = 1/2 x I x W^2 + 1/2 x m x ( r x w)^2

solve for w , w = 12.27 rad/sec (ANSWER D)

speed = rx w = .7x 12.27 = 8.5 m/sec (ANSWER D)

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