Question

A thin, uniform rod is hinged at its midpoint. To begin with, one-half of the rod is bent upward and is perpendicular to the other half. This bent object is rotating at an angular velocity of 9.7 rad/s about an axis that is perpendicular to the left end of the rod and parallel to the rod's upward half (see the drawing). Without the aid of external torques, the rod suddenly assumes its straight shape. What is the angular velocity of the straight rod?

Answer 1

Answer:

The angular velocity of the straight rod is 4.85 rad/s.

Explanation:

Given that,

Initial angular velocity =9.7 rad/s

We need to calculate the initial moment of inertia of the rod

Using formula of moment of inertia

$I_{i}=\dfrac{1}{3}(\dfrac{M}{2})r^2+(\dfrac{M}{2})r^2$

$I_{i}=\dfrac{2}{3}Mr^2$

We need to calculate the final moment of inertia of the rod

Using formula of moment of inertia

$I_{f}=\dfrac{1}{3}M(2r)^2$

$I_{f}=\dfrac{4}{3}Mr^2$

We need to calculate the angular velocity of the straight rod

Using conservation of rotational kinetic energy

$I_{i}\omega_{i}=I_{f}\omega_{f}$

Put the value into the formula

$\dfrac{2}{3}Mr^2\times9.7=\dfrac{4}{3}Mr^2\times\omega_{f}$

$\omega_{f}=\dfrac{2\times9.7}{4}$

$\omega_{f}=4.85\ rad/s$

Hence, The angular velocity of the straight rod is 4.85 rad/s.

Answer 2

Answer:

$\omega_1 = 6.06 rad/s$

Explanation:

Initially rod is bent into L shaped at mid point

So we will have rotational inertia of the rod given as

$I = \frac{mL^2}{3} + (\frac{mL^2}{12} + m(L^2 + \frac{L^2}{4}))$

$I = \frac{5mL^2}{3}$

Now when rod becomes straight during the rotation

then we will have

$I_2= \frac{(2m)(2L)^2}{3} = \frac{8mL^2}{3}$

now from angular momentum conservation we have

$I \omega_o = I_1 \omega_1$

$(\frac{5mL^2}{3}) (9.7) = \frac{8mL^2}{3}\omega_1$

$\omega_1 = 6.06 rad/s$