Question

A rod of mass m and length 2R can rotate about an axis passing through O in vertical plane. A disc of mass m and radius R is hinged to the other end P of the rod and can freely rotate about P. When disc is at lowest point both rod and disc has angular velocity w. If rod rotates by maximum angle 60o with downward vertical, then find w in terms of R and g. (all hinges are smooth) {Ans: (9g/16R)1/2}

Answer 1

The angular velocity is $\sqrt{\dfrac{9g}{16R}}$

Explanation:

Given that,

Mass of rod = m

Length of rod = 2R

Mass of disc = m

Length of disc = R

We need to calculate the moment of inertia of the rod

Using formula of moment of inertia

$I=\dfrac{1}{3}mr^2$

Put the value into the formula

$I=\dfrac{1}{3}m(2R)^2$

We need to calculate the moment of inertia of the disc

Using formula of moment of inertia

$I=\dfrac{1}{2}mr^2+4mr^2$

Put the value into the formula

$I=\dfrac{1}{2}mR^2+4mR^2$

$I=\dfrac{9mR^2}{2}$

Loss in potential energy by the rod and the disc

$P.E=mgh+mgh'$

Put the value of angle

$P.E=mgR(1-\cos60)+2mgR(1-\cos60)$

$P.E=\dfrac{3mgR}{2}$

We need to calculate the angular velocity

Potential energy converted in kinetic energy

$P.E=K.E$

$\dfrac{3mgR}{2}=\dfrac{1}{2}I\omega^2+\dfrac{1}{2}mv^2$

Put the value of v

$\dfrac{3mgR}{2}=\dfrac{1}{2}(\dfrac{1}{3}m(2R)^2+\dfrac{1}{2}m(2R\omega)^2$

$\dfrac{3gR}{2}=(\dfrac{2R^2}{3}+2R^2)\omega^2$

$\dfrac{3gR}{2}=(\dfrac{8R^2}{3})\omega^2$

$\omega^2=\dfrac{9g}{16R}$

$\omega=\sqrt{\dfrac{9g}{16R}}$

Hence, The angular velocity is $\sqrt{\dfrac{9g}{16R}}$

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Topic : angular velocity

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