A 2.20 kg hoop 2.00 m in diameter is rolling to the right without slipping on a

Question

A 2.20 kg hoop 2.00 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.40{ m rad/s} .

1-How fast is its center moving?

2-What is the total kinetic energy of the hoop?

3-Find the magnitude of the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop, (iii) a point on the right side of the hoop, midway between the top and the bottom.

4-Find the magnitude of the velocity vector for each of the points in part (C), except as viewed by someone moving along with same velocity as the hoop.

Explanation / Answer

velocity at the center =r*w=1*2.4=2.4 m/sec total kinectic energy=0.5mv^2+0.5Iw^2=0.5(2.2*2.4^2+0.5*2.2*1^2*2.4^2)=9.504joules i) at velocity at heighest point rw+rw=2rw=2*2.4*1=4.8m/sec in x direction ii) at lowest point zero pure rolling iii)at right side point=sqrt( velocity at center ^2+(rw)^2)=sqrt(2)*2.4=3.39m/sec if move along with hoop it is vector summ of two direction 45degrss downward with x axis i)rw=2.4m/sec x iderction ii)zero iii)rw=2.4 m/sec in -ve y direction

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