Question

A hoop and a disk, both of 0.88- m radius and 4.0- kg mass, are released from the top of an inclined plane 3.3 m high and 8.1 m long. What is the speed of each when it reaches the bottom? Assume that they both roll without slipping. What is the speed of the hoop? What is the speed of the disk?

Answer 1

Using energy conservation

FOR HOOP

KEi + PEi = KEf + PEf

PEf = 0, at ground

KEi = 0, intially speed is zero

PEi = KEf

PEi = KEtrans. + KErot

m*g*h = 0.5*m*V^2 + 0.5*I*w^2

w = V/r

I = moment of inertia of hoop = m*r^2

m*g*h = 0.5*m*V^2 + 0.5*(m*r^2)*(v/r)^2

g*h = V^2/2 + V^2/2

g*h = V^2

V = sqrt (g*h)

Given that h = 3.3 m

V = sqrt (9.81*3.3)

V = 5.7 m/sec = speed of hoop at the bottom

FOR DISK

KEi + PEi = KEf + PEf

PEf = 0, at ground

KEi = 0, intially speed is zero

PEi = KEf

PEi = KEtrans. + KErot

m*g*h = 0.5*m*V^2 + 0.5*I*w^2

w = V/r

I = moment of inertia of disk = m*r^2/2

m*g*h = 0.5*m*V^2 + 0.5*(m*r^2/2)*(v/r)^2

g*h = V^2/2 + V^2/4

g*h = 3*V^2/4

V = sqrt (4*g*h/3)

Given that h = 3.3 m

V = sqrt (4*9.81*3.3/3)

V = 6.6 m/sec = speed of disk at the bottom

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