A hoop and a disk, both of 0.66- m radius and 4.0- kg mass,are released from the

Question

A hoop and a disk, both of 0.66- m radius and 4.0- kg mass,are released from the top of an inclined plane 3.1 m high and 8.9 mlong. What is the speed of each when it reaches the bottom? Assumethat they both roll without slipping.
a) What is the speed of the hoop?
b) What is the speed of the disk? A hoop and a disk, both of 0.66- m radius and 4.0- kg mass,are released from the top of an inclined plane 3.1 m high and 8.9 mlong. What is the speed of each when it reaches the bottom? Assumethat they both roll without slipping.
a) What is the speed of the hoop?

Explanation / Answer

The moment of inertia of ahoop=mr2=4*0.662=1.74 kgm2                                   of a disk=1/2mr2=0.87 kgm2 At the top of the incline, both have only potential energy of valuemgh =4*9.8*3.1=121.52J All this energy is converted into kinetic and rolling energy. PE=1/2mv2 + 1/2I2 (I:moment ofinertia, :angular velocity) since the body rolls without slipping, v=r, =v/r For the hoop: 121.52J=1/2*4*v2 +1/2*1.74*v2/0.662             =2v2 +1.99v2 approx 4v2=121.52J v=5.52m/s for the disk: 121.52J=2v2 + 1v2 3v2=121.52 v=6.36 m/s

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