System of Particles and Rotational Motion Test - 15

Yes, when a body has a uniform mass density, its centre of mass shall coincide with its geometrical centre.

This is because entire mass of ring is at its periphery i.e. at maximum distance from the centre. The mass of disc is distributed from the centre to the ring.

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 \(\theta\)) x s = 280

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

= 57.14 m

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

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

\(\frac{1}{2}mv^2 + \frac{1}{2}I \frac{v^2}{r^2}\) = mg\(\Big(\frac{3v^2}{4g} \Big)\)  \(\Big(\because \omega = \frac{v}{r} \Big)\)

\(\frac{1}{2}I\frac{v^2}{r^2} = \frac{3}{4}mv^2 - \frac{1}{2}mv^2 = \frac{1}{4}mv^2\)

or 

\(I = \frac{1}{2}mr^2 \; ,as (\frac{1}{2} mr^2)\)

is a moment of inertia of a circular disc of mass m and radius r about an axis passing through its centre and perpendicular to its plane, the object must be a circular disc.

It lies at the centroid of the triangular lamina i.e. where the three medians of the triangular lamina intersect.

An isolated system is that on which no external force is acting.

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

When a disc rotates with uniform angular velocity, angular acceleration of the disc is zero.

As there is no external torque as jump is radially away.

Applying conservation of angular momentum

\(I_1\omega_1 = I_2 \omega_2\)

\(I_1\) = Initial moment of inertia = \(mr^2 + mr^2 = 2mr^2\)

\(I_2\)  = Final moment of inertia

ω = Initial angular velocity

As the person jumps with angular velocity ω. Hence its angular momentum at that time is I ω. So final angular momentum is I ω for the person.

So,

\(2mr^2\omega = mr^2\omega_1 + mr^2\omega\)

\(\omega_1 = \omega\)

For a general rotational motion angular momentum L and angular velocity ω are not parallel when the axis of rotation is not symmetric As L = I ω and I is not a scalar quantity.

Whereas in translational motion, momentum vector (p) = m vector (v), where m is scalar so direction of p and v are always same.

As we know that

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

When forces are acting radially then θ = 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

Centre of mass of body remains unchanged in rotational motion of rigid body.

Rot. K.E. = \(\frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{2}{5} MR^2 \times \frac{v^2}{R^2}\)

\(\Big(as \;\omega = \frac{v}{R}, I = \frac{2}{5}MR^2 \Big)\)

= \(\frac{1}{5}mv^2\)

Total energy = Translational K.E. + Rot.K.E.

\(= \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2\)

\(\therefore \frac{Rot.\; K.E.}{Total\;Energy} = \frac{\frac{1}{5}mv^2}{\frac{7}{10} mv^2} = \frac{2}{7}\)

Rotational K.E = (\(\frac{1}{2}\)) \(I \omega^2\)

 Hence (\(\frac{K_1}{K_2}\)) = (\(\frac{I_1}{I_2}\)) \((\frac{\omega_1}{\omega_2})^2 \)

Given \(K_1\) = K, \(\omega_1\) = ω also \(\omega_2\) = ω/ \(n^2\)

 Hence \((\frac{K}{K_2}) = (\frac{I_1}{I_2})(\frac{1}{n^2})^2\)

\(\therefore (\frac{K}{K_2}) = \frac{\frac{2}{5}MR_1^2}{\frac{2}{5}M(\frac{R_1}{n})^2} \times {(\frac{1}{n^2})^2}\)

\(\frac{K}{K_2} = \frac{n^2}{n^4} = \frac{1}{n^2}\)

\(K_2 = n^2.K\)

Moment of Inertial of a solid sphere of radius r and mass m about its diameter is given as

\(I_1 =\frac{2}{5}m \times r^2\)

Now if radius of the sphere is doubled keeping mass constant then,

\(I_2 = \frac{2}{5}m \times (2r)^2\)

\(\therefore \frac{I_1}{I_2} = \frac{1}{4}\)

The tangent is parallel to its radius. The distance between two axes is R. Using theorem of parallel axes.

\(I_{tangent} = I_{diameter} + IR^2\)

= \(\frac{MR^2}{2} + MR^2 \)

= \(\frac{3}{2}MR^2\)

The linear K.E of sphere is simply \(\frac{mv^2}{2}\)

Rotational K.E of a sphere is \(\frac{I\omega^2}{2}\)

If there is no sliding, slipping then \(\omega = \frac{v}{r}\)

If the solid sphere has a uniform density then \(I=\frac{2mr^2}{5}\)

If the two assumption's above hold then rotational K.E of sphere 

\(=\frac{\left(\frac{2mr^2}{5}\right)\left(\frac{v}{r}\right)^2}{2}\)

\(=\frac{mv^2}{5}\)

Ratio \(=\frac{\frac{mv^2}{5}}{\frac{mv^2}{2}}\) \(=\frac{2}{5}\)

When a dancer is spinning, she has a certain angular momentum.

When the dancer folds her arms, the distance of all the points of her body decreases with respect to axis of rotation. Thus, in order to conserve angular momentum, the speed of rotation has to increase and hence the dancer spins faster.

 Similarly, when the dancer stretches her arms, the distance of all the points of her body increases with respect to axis of rotation. Thus, in order to conserve angular momentum, the speed of rotation has to decrease and hence the dancer spins slower.

Angular momentum of particle w.r.t origin = linear momentum x perpendicular distance of line of action of linear momentum from origin.

= (mv) x (a)

= mva = constant