A 2.90kg hoop 1.50m in diameter is rolling to the right without slipping on a ho

Question

A 2.90kg hoop 1.50m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.10rad/s.

A.)How fast is its center moving? (Answer in m/s)

B.)What is the total kinetic energy of the hoop? (Answer in J)

C.)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. (Answers in m/s)

D.) To be continued

E.) 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. (Vi, Vii, Viii in m/s)

F.) To be continued

Explanation / Answer

1. v = r*? = 0.75m*2.10 rad/s = 1.575m/s

2. K = 1/2*m*v^2 + 1/2*I*?^2 = 1/2*2.90*1.575^2 + 1/2*(2.90*0.75^2)*2.100^2 = 7.19 J

3.i) v = vtr + vr = 1.575 + 1.575 = m/s where vtr is the translational velocity and vr is the rotational velocity

ii) v = vtr + vr =1.575 - 1.575 = 0 m/s

iii) Here the velocities must be added as vectors so v = sqrt(vtr^2 + vr^2) = sqrt(1.575^2 + 1.575^2) =

2.22m/s

4.Now vtr is zero s seen by an observer moving with the loop

so v at all three is the rotational velocity so v = 1.575m/s

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