Question

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the string is held in place and the hoop is released from rest (Fig. E10.20). After the hoop has descended 75.0 cm, calculate (a) the angular speed of the rotating hoop and (b) the speed of its center.

Answer 1

Answer

given,

radius of the hoop = 8 cm = 0.08 m

mass = 0.18 kg

hoop descended to 75 cm = 0.75 m

conservation of energy

$\dfrac{1}{2}I_1\omega_1^2 + \dfrac{1}{2}mv_1^2 + mgh_1 = \dfrac{1}{2}I_2\omega_2^2 + \dfrac{1}{2}mv_2^2 + mgh_2$

here $\omega_1 = v_1 = 0$

$mgh_1= \dfrac{1}{2}I_2\omega_2^2 + \dfrac{1}{2}mv_2^2 + mgh_2$

$mg(h_2-h_1)= \dfrac{1}{2}I_2\omega_2^2 + \dfrac{1}{2}mv_2^2$

$h_2 - h_1 = h$

$mgh= \dfrac{1}{2}I_2\omega_2^2 + \dfrac{1}{2}mv_2^2$

$mgh= \dfrac{1}{2}mr^2\omega_2^2 + \dfrac{1}{2}m(r\omega_2)^2$

$r^2\omega_2^2 = gh$

$\omega_2 = \dfrac{\sqrt{gh}}{r}$

$\omega_2 = \dfrac{\sqrt{9.8\times 0.75}}{0.08}$

$\omega_2 = 33.89\ rad/s$

b) $v = r \omega$

       = 33.89 x 0.08

    v = 2.704 m/s

Answer 2

Answer:

(a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

Explanation:

Given that,

Radius = 8.00 cm

Mass = 0.180 kg

Height = 75.0 m

We need to calculate the angular speed of the rotating

Using conservation of energy

$\dfrac{1}{2}I\omega_{1}^2+\dfrac{1}{2}mv_{1}^{2}+mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

Here, initial velocity and angular velocity are equal to zero.

$mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

$mg(h_{1}-h_{2})=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}$

$mgH=\dfrac{1}{2}mr^2\omega_{2}^2+\dfrac{1}{2}m(r\omega_{2})^2$

Here, $H = h_{1}-h_{2}$

$gH=r^2\omega_{2}^2$

$\omega_{2}^2=\dfrac{gH}{r^2}$

$\omega_{2}=\sqrt{\dfrac{9.8\times75.0\times10^{-2}}{(8.00\times10^{-2})^2}}$

$\omega_{2}=33.8\ rad/s$

The angular speed of the rotating is 33.8 rad/s.

(b). We need to calculate the speed of its center

Using formula of speed

$v=r\omega$

Put the value into the formula

$v=8.00\times10^{-2}\times33.8$

$v=2.7\ m/s$

Hence, (a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.