A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2 5 times the initial value? Assume that the turntable rotates without friction. Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Given that the initial angular speed of the turntable, ω i = 40 rev/min Let the initial moment of inertia of the system is I When the child folds his hands, system moment inertia becomes, 2 5 I From the consecration of angular momentum, Initial total angular momentum = Final total angular momentum I × 40 = 2 5 I × ω f ⇒ ω f = 100 rev/min The angular speed of the system after the child folding his hands is 100 rev/min Now, Kinetic energy, K E = L 2 2 I As the angular momentum is conserved, Final kinetic energy = 5 2 Initial kinetic energy The child uses his internal energy to increase the kinetic energy.

A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2 5 times the initial value? Assume that the turntable rotates without friction. Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Given that the initial angular speed of the turntable, ω i = 40 rev/min Let the initial moment of inertia of the system is I When the child folds his hands, system moment inertia becomes, 2 5 I From the consecration of angular momentum, Initial total angular momentum = Final total angular momentum I × 40 = 2 5 I × ω f ⇒ ω f = 100 rev/min The angular speed of the system after the child folding his hands is 100 rev/min Now, Kinetic energy, K E = L 2 2 I As the angular momentum is conserved, Final kinetic energy = 5 2 Initial kinetic energy The child uses his internal energy to increase the kinetic energy.

E M B I B E

PHYSICS PART 1 TEXTBOOK FOR CLASS XI>Chapter -1 - Systems of Particles and Rotational Motion>EXERCISES>Q 7.13

EASY

11th CBSE

IMPORTANT

Earn 100

A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of $40$ rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $\frac{2}{5}$ times the initial value? Assume that the turntable rotates without friction. Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Important Points to Remember in Chapter -1 - Systems of Particles and Rotational Motion from NCERT PHYSICS PART 1 TEXTBOOK FOR CLASS XI Solutions

1. Centre of mass of a system of n discrete particles:

${\vec{r}}_{c m} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2} + \dots \dots \dots + m_{n} {\vec{r}}_{n}}{m_{1} + m_{2} + \dots \dots . . + m_{n}}$

2. Centre of mass of a continuous mass distribution:

$x_{c m} = \frac{\int x d m}{\int d m}, y_{c m} = \frac{\int y d m}{\int d m}, z_{c m} = \frac{\int z d m}{\int d m}$

3. Velocity of centre of mass of system:

(i) ${\vec{v}}_{c m} = \frac{m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2} + \dots \dots \dots + m_{n} {\vec{v}}_{n}}{M}$

(ii) Momentum of a system given by, ${\vec{P}}_{S y s t e m} = M {\vec{v}}_{c m}$

4. Acceleration of centre of mass of system:

(i) ${\vec{a}}_{c m} = \frac{m_{1} {\vec{a}}_{1} + m_{2} {\vec{a}}_{2} + \dots \dots \dots + m_{n} {\vec{a}}_{n}}{M}$
(ii) The net force acting on a system is given by, ${\vec{F}}_{e x t} = M {\vec{a}}_{c m}$

5. Cross product:

(i) Cross product of two vectors $\vec{A}$ and $\vec{B}$ can be written as $\vec{A} \times \vec{B} = A B \sin θ \hat{n}$ , where $θ$ is the angle between the vectors, $\hat{n}$ is unit vector perpendicular to both $\vec{A}$ and $\vec{B}$ .

(ii) If $\vec{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$ and $\vec{B} = B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k}$ then, $\vec{A} \times \vec{B} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{matrix}|$

6. Moment of Inertia:

(i) For a single particle, $I = m r^{2}$ where $m$ is the mass of the particle and $r$ is the perpendicular distance of the particle from the axis about which moment of Inertia is to be calculated.

(ii) Moment of inertia of a system of particles: $I = m r_{1}^{2} + m_{2} r_{2}^{2} + . . .$ $= I_{1} + I_{2} + I_{3} + . . . .$

(iii) Moment of inertia of a continuous object: $I = \int d m r^{2}$ , where $d m$ is the mass of a small element, $r$ is the perpendicular distance of the element from the axis

7. Perpendicular Axis Theorem (Only applicable to plane lamina):

$I_{z} = I_{x} + I_{y}$ (when object is in $x - y$ plane).

8. Parallel Axis Theorem (Applicable to any type of object):

$I = I_{c m} + M d^{2}$ , here $I_{c m}$ is the moment of inertia about an axis passing through centre of mass and parallel to required axis and $d$ is the perpendicular distance between required axis and axis passing through centre of mass.

9. Moment of inertia of few structures:

(i) Solid sphere: $\frac{2}{5} M R^{2}$

(ii) Hollow sphere: $\frac{2}{3} M R^{2}$

(iii) Ring: $M R^{2}$

(iv) Disc: $\frac{1}{2} M R^{2}$

(v) Hollow cylinder: $M R^{2}$

(vi) Solid cylinder: $\frac{1}{2} M R^{2}$

(vii) Rod about centre: $\frac{M L^{2}}{12}$

(viii) Rectangular plate: $\frac{M (a^{2} + b^{2})}{12}$

10. Moment of inertia ( $I$ ) in terms of radius of gyration $K$ :

$I = M K^{2}$

11. Torque:

(i) $\vec{τ} = \vec{r} \times \vec{F}$

(ii) $|τ| = r F \sin θ = r_{⊥} F = F_{⊥} r$

12. Angular momentum $(\vec{L})$ of a particle about a point:

(i) $\vec{L} = \vec{r} \times \vec{P}$

(ii) $|L| = r P \sin θ = r_{⊥} P = P_{⊥} r$

13. Angular momentum of a rigid body:

(i) $L_{H} = I_{H} ω$ , where $L_{H} =$ angular momentum of object about axis $H$ . $I_{H} =$ Moment of Inertia of rigid object about axis $H$ . $ω =$ angular velocity of the object.

(ii) ${\vec{L}}_{A B} = I_{c m} \vec{ω} + {\vec{r}}_{c m} \times (M {\vec{v}}_{c m})$

14. Conservation of angular momentum:

According to conservation of angular momentum, if $τ_{e x t} = 0$ about a point or axis of rotation, angular momentum of the particle or the system remains constant about that axis or point.

15. Relation between Torque and Angular Momentum:

$\vec{τ} = \frac{d \vec{L}}{d t}$

16. Kinetic energy:

Total K.E. $= \frac{1}{2} M v_{c m}^{2} + \frac{1}{2} I_{c m} ω^{2}$