A thin hoop of mass m and radius L rolls without slipping as shown in the figure

Question

A thin hoop of mass m and radius L rolls without slipping as shown in the figure. It is supported by an axle of negligible mass and length R through its center. The axle rotates about the Y-axis with a constant angular velocity %u03A9. The moment of inertia of the hoop about the axle which passes through its center of mass is I.

(a)What is the angular momentum of the hoop as a function of %u03B8?

(b)What is the kinetic energy of the hoop?

Your answer can contain the symbols m, I, %u03A9, L, %u03B8, and R. For this problem assume that the center of mass of the hoop coincides with its geometric center.

Explanation / Answer

No slip condition:

R*OMEGA = L*w.....where w = angular valocity of hoop about its own axis.

Hence, w = R*OMEGA / L

There are three contributions to the angular momentum: the

horizontal component (often known as the %u201Cspin%u201D angular momentum), the motion of the

center of the wheel about the central shaft (often known as the %u201Corbital%u201D angular

momentum) and the fact that the wheel is also rotating about a vertical axis. The angular

momentum is then given by

L = w*mR^2 (-r_hat) + OMEGA*mR^2 (j_hat) + OMEGA*I (j_hat) ...........where j_hat = unit vector in y-direction and r_hat = unit vector in radially outward directinon.

Thus, L = (R*OMEGA/L)*mR^2 (-r_hat) + OMEGA*mR^2 (j_hat) + OMEGA*I (j_hat)

L = -OMEGA*mR^3 /L r_hat + OMEGA*(I + mR^2) j_hat

r_hat = Cos theta k_hat + Sin theta i_hat.........where i_hat = unit vector in x-direction and k_hat = unitvector in z-direction

Thus, L = -OMEGA*mR^3 /L (Cos theta k_hat + Sin theta i_hat) + OMEGA*(I + mR^2) j_hat

b)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.