Question

A bead of mass m slides without friction on a vertical hoop of radius R . The bead moves under the combined action of gravity and a spring, with spring constant k , attached to the bottom of the hoop. Assume that the equilibrium (relaxed) length of the spring is R. The bead is released from rest at θ = 0 with a non-zero but negligible speed to the right.
(a) What is the speed v of the bead when θ = 90∘ ? Express your answer in terms of m, R, k, and g.
(b) What is the magnitude of the force the hoop exerts on the bead when θ = 90∘ ? Express your answer in terms of m, R, k, and g.

Answer 1

A bead of mass m slides without friction on a vertical hoop of radius R. The bead moves under the combined action of gravity and a spring, with spring constant k, attached to the bottom of the hoop. Assume that the equilibrium (relaxed) length of the spring is R. The bead is released from rest at $\theta = 0$ with a non-zero but negligible speed to the right.

(a) What is the speed v of the bead when $\theta = 90{}^{\circ}$? Express your answer in terms of m, R, k, and g.

(b) What is the magnitude of the force the hoop exerts on the bead when $\theta = 90{}^{\circ}$? Express your answer in terms of m, R, k, and g.

Solution

As the equilibrium (relaxed) length of the spring is R the force due to the spring is $k(r - R)$, where r is the length of the spring.

At the top of the hoop the gravitational potential energy of the bead is $mg(2R)$ and potential energy due to the spring is $\frac{1}{2} k(2R - R)^2 = \frac{1}{2} kR^2$.

Hence the initial potential energy is

$$U_i = \frac{1}{2} k R^2 + 2 m g R.$$

Since all the forces are conservative, the mechanical energy is constant and we have

$$K_i + U_i = K_f + U_f.$$

The initial kinetic energy is zero and we obtain

$$K_f = U_i - U_f.$$

When $\theta = 90{}^{\circ}$ the gravitational potential energy of the bead is $mgR$. The length of the spring is $\sqrt{R^2 + R^2} = \sqrt{2} R$. Potential energy due to the spring is $\frac{1}{2} k\left(\sqrt{2} R - R\right)^2 = \frac{1}{2} k R\left(\sqrt{2} - 1\right)^2$.

Hence the final potential energy is

$$U_f = \frac{1}{2} k R^2 \left(\sqrt{2} - 1\right)^2 + m g R.$$

We have

$$\frac{m v_f^2}{2} = \left(\frac{1}{2} k R^2 + 2 m g R\right) - \left(\frac{1}{2} k R^2 \left(\sqrt{2} - 1\right)^2 + m g R\right).$$

The speed of the bead when $\theta = 90{}^{\circ}$

$$v_f = \sqrt{\frac{2(\sqrt{2} - 1) k R^2}{m} + 2 g R}$$

We can apply the second Newton's law to the bead when $\theta = 90{}^{\circ}$:

$$\overrightarrow {m a} = \overrightarrow {m g} + \overline {{k (\sqrt {2} R - R)}} + \overrightarrow {N}.$$

Consider radial projection of it:

$$0 = k \left(\sqrt {2} R - R\right) \cos 4 5 {}^ {\circ} - N.$$

The magnitude of the force the hoop exerts on the bead when $\theta = 90{}^{\circ}$

$$N = k \big (\sqrt {2} R - R \big) \cos 4 5 {}^ {\circ} = k R \big (\sqrt {2} - 1 \big) \frac {\sqrt {2}}{2} = k R \left(1 - \frac {1}{\sqrt {2}}\right).$$

Answer: (a) $\sqrt{\frac{2(\sqrt{2} - 1)kR^2}{m} + 2gR}$ ; (b) $kR\left(1 - \frac{1}{\sqrt{2}}\right)$ .