System of Particles and Rotational Motion Test - 16

Here the velocity of center of mass has been asked so we need to consider only the translational motion so

\(v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 7}\)

= 11.71 m/s

\(a_c = \frac{g sin \theta}{1 + \frac{K^2}{R^2}}\)

\(a_c = \frac{10 sin 30}{1 + \frac{K^2}{R^2}}\)

\(a_c = \frac{10 \times 0.5}{1 + \frac{1}{2}}\)

As sin 30 = \(\frac{1}{2} = 0.5\)

Now, \(a_c = \frac{5}{\frac{3}{2}}\)

\(a_c = \frac{5 \times 2}{3}\)

\(a_c = \frac{10}{3}\)

(i) Assuming these cylinders are identical in every respect with the exception of one being solid and the other hollow, and also if conditions on the slope are equal and with no interference from wind, etc then the cylinders should roll exactly the same so long as the hollow one is heavy enough.

(ii) If the hollow cylinder is made of a very light material then drag may play a factor in slowing it down.

\(\mathrm{K.E_{Translatory}=K.E_{Rotatary}}\)

\(\mathrm{\therefore\,\frac{1}{2}mv^2=\frac{1}{2}mv^2\left(\frac{k^2}{R^2}\right)}\)

\(\mathrm{\Rightarrow\frac{K^2}{R^2}=1}\)

This value of \(\mathrm{\frac{K^2}{R^2}}\) confirms the body as hollow cylinder.

For ring, \(\frac{k^2}{R^2} = 1\)

Solid sphere =\(\frac{k^2}{R^2} = \) \(\frac{2}{5} = 0.4\)

\(Disc, \frac{k^2}{R^2} = 0.5\)

When the solid cylinder rolls down, its potential energy is converted to its kinetic energy (rotational & translational).

\(mgh = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2\)

\(gh = \frac{1}{2} (r \omega)^2 + \frac{1}{2} (\frac{1}{2} r^2) \omega^2\)

\(gh = \frac{3}{4} (r\omega)^2\)

\(gh = \frac{3}{4} v^2\)

\(v = \sqrt{\frac{4gh}{3}}\)

Substitute values,

\(v = \sqrt{4 \times 9.8 \times \frac{2}{3}}\)

v = 5.11 m/s

Therefore, linear speed is 5.11 m/s.

Since the inclined plane is frictionless, then there will be no rolling and the mass will only slide down

Hence acceleration a = gsin\(\theta\) is same for solid sphere, hollow sphere, and ring.

\(\frac{\frac{1}{2} I \omega^2}{\frac{1}{2} mv^2} = \frac{40}{60}\)

\(v = R \omega\)

Hence, \(I = \frac{2}{3} m R^2\)

\(KE_{rot} = \frac{1}{2}I \omega^2\)

but \(v = \omega R\)

or, \(\omega = \frac{v}{R}\)

And the moment of inertia for a hollow sphere is I = \(\frac{2}{3}MR^2\)

\(\therefore KE_{rot} = \frac{1}{2}\frac{2}{3}m R^2(\frac{v}{R})^2\)

Translational KE:

\(KE_{tr} = \frac{1}{2}mv^2\)

Total KE:

\(KE_{total} = KE_{rot} + K_{Etr}\)

\(KE_{total} = \frac{1}{3}mv^2+\frac{1}{2}mv^2=\frac{5}{6}mv^2\)

The ratio of the rotational KE to the total KE is:

\(\frac{KE_{rot}}{KE_{total}} = \frac{\frac{1}{3}mv^2}{\frac{5}{6}mv^2} = \frac{2}{5}\)

For solid sphere, I = \(\frac{2}{5}MR^2\)

Also, v = \(R \omega\) (no slipping)

Rotational K.E, \(E_R = \frac{1}{2}I \omega^2 = \frac{1}{2} \times \frac{2}{5} MR^2 \omega^2\) = \(\frac{1}{5}Mv^2\)

Total K.E

E = \(\frac{1}{2}I\omega^2 + \frac{1}{2} Mv^2 = \frac{7}{10}Mv^2\) 

\(\frac{E_R}{E} = \frac{2}{7}\)

Moment of inertia of a ring about its diameter I = \(\frac{1}{2}mr^2 \) and kinetic energy is given by

\(E_k = \frac{1}{2}I\omega^2\) ⇒ \(E_k = \frac{1}{4}mr^2 \omega^2\)

= \(\frac{1}{4} \times 10 \times (0.5)^2 \times (20)^2 = 250 J\)

By pulling his arms and legs in, decreases his moment of inertia, thereby increasing his angular velocity using the principle of conservation of angular momentum.

The term moment of momentum is called Angular Momentum.

The kinetic energy of the rolling object is converted into potential energy at height h

\(\frac{7}{10}mv^2 = \frac{7}{10} \times 1 \times (20)^2 = 280J\)

\((mg \;sin \theta) \times h = 280\)

s = \(\frac{280}{mg\;sin\theta} = \frac{280}{1\times 9.8\times \frac{1}{2}}\)

= 57.14 m

The centre of mass of the bangle (circular in shape) lies at its centre.

As we know that

\(\tau = \vec r \times \vec F = rFsin \theta\)

When forces are acting radially then \(\theta\) = 0. So, \(\tau\) = 0

When forces are acting on axis of rotation then r = 0. So, \(\tau\) = 0

When forces are acting parallel to axis of rotation then its component in plane r and F is 0. Hence F = 0. So, \(\tau\) = 0

\(\mathrm{K=\frac{1}{2} I\omega^2=\frac{1}{2}\big(\frac{2}{5} MR^2\big).\omega^2}\)

\(\mathrm{K'=\frac{1}{2}I'\omega' =\frac{1}{2}\big(\frac{2}{5} \frac{MR^2}{n^2}\big).\omega'^2}\)

\(\mathrm{\frac{K'}{K}=\frac{1}{n^2}.\frac{\omega'^2}{\omega^2}}\)

\(\mathrm{=\frac{1}{n^2}.(n^2)^2}\)

\(\mathrm{=n^2}\)

\(\mathrm{K'=n^2K}\)

\(\mathrm{\frac{K'}{K}=n^2}\)

Given moment of inertia of the wheel about its axis is \(160kgm^2\)

I = \(MK^2\) (K is the radius of gyration)

160 = \(10 \times K^2\)

K = 4m

Radius of gyration of the wheel about its own axis is 4 m.

(i) The moment of inertia will decrease as distance of the mass from the axis of rotation will decrease.

(ii) The moment of inertia of the sphere will be less than that of the ring.