a) A ballet dancer sometimes stretches and sometimes folds her arms during her performance, why? Justify. [2] b) Establish a relation between torque and moment of inertia for a rigid body. [2] c)…

Model Answer

a) A ballet dancer sometimes stretches and sometimes folds her arms during her performance, why? Justify.

A ballet dancer changes her moment of inertia by stretching or folding her arms to conserve angular momentum during rotations. This principle is based on the law of conservation of angular momentum, which states that if no external torque acts on a system, its angular momentum remains constant.

• When a dancer stretches her arms, her mass distribution moves farther from the axis of rotation, increasing her moment of inertia (I).
  • Since angular momentum $L=I\omega$ must remain constant, her angular velocity $\omega$ decreases, slowing her rotation.
• When she folds her arms, her moment of inertia decreases, causing her angular velocity to increase, making her spin faster.

This control over rotation speed is crucial for ballet spins, jumps, and graceful movements.

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b) Establish a relation between torque and moment of inertia for a rigid body.

For a rigid body rotating about a fixed axis, torque (τ) is related to angular acceleration (α) and moment of inertia (I) by Newton’s second law for rotational motion:

$$\tau =I\alpha$$

Explanation:

• Torque (τ) is the rotational equivalent of force and causes angular acceleration.
• Moment of inertia (I) is the rotational equivalent of mass, representing an object’s resistance to rotational motion.
• Angular acceleration (α) is the rate of change of angular velocity.

Thus, the greater the torque applied, the greater the angular acceleration, provided the moment of inertia remains constant. Conversely, for a given torque, a larger moment of inertia results in a smaller angular acceleration.

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c) Why do we prefer a wrench of longer arm over a wrench of shorter arm?

A longer wrench provides a greater torque (τ) for the same applied force (F), due to the increased lever arm (r) in the equation:

$$\tau =r\times F$$

• A longer arm increases the moment arm (r), making it easier to loosen or tighten bolts with less effort.
• This is particularly useful when dealing with tight or stubborn fasteners.

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OR

a) Show that the total energy of the particles executing SHM is constant.

In Simple Harmonic Motion (SHM), the total mechanical energy (E) is the sum of kinetic energy (KE) and potential energy (PE), and it remains constant if there is no damping.

Derivation:

1. Displacement in SHM: $x=A\sin (\omega t+\varphi )$
2. Velocity in SHM: $v=\frac{dx}{dt}=A\omega \cos (\omega t+\varphi )$
3. Kinetic Energy (KE): $$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m(A\omega \cos (\omega t+\varphi ){)}^2=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t+\varphi )$$
4. Potential Energy (PE) in SHM: $$PE=\frac{1}{2}k{x}^2=\frac{1}{2}k(A\sin (\omega t+\varphi ){)}^2=\frac{1}{2}k{A}^2{\sin }^2(\omega t+\varphi )$$ Since $k=m{\omega }^2$, we have: $$PE=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2(\omega t+\varphi )$$
5. Total Energy (E): $$E=KE+PE=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t+\varphi )+\frac{1}{2}m{A}^2{\omega }^2{\sin }^2(\omega t+\varphi )$$ $$E=\frac{1}{2}m{A}^2{\omega }^2({\cos }^2(\omega t+\varphi )+{\sin }^2(\omega t+\varphi ))=\frac{1}{2}m{A}^2{\omega }^2$$ Since ${\cos }^2\theta +{\sin }^2\theta =1$, the total energy remains constant throughout the motion.

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b) Draw a graph showing the variation of potential energy of a particle in SHM with displacement.

Explanation:

• The graph shows a parabolic curve because $PE=\frac{1}{2}k{x}^2$.
• At equilibrium (x = 0), PE is minimum (zero).
• At maximum displacement (x = ±A), PE is maximum ($\frac{1}{2}k{A}^2$).

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c) The time period of a simple pendulum on the surface of Earth is T. What will be its time period on the Moon’s surface?

The time period (T) of a simple pendulum is given by:

$$T=2\pi \sqrt{\frac{L}{g}}$$

• On Earth, $T=2\pi \sqrt{\frac{L}{g_E}}$
• On the Moon, $g_M=\frac{g_E}{6}$ (since the Moon’s gravity is $\frac{1}{6}$th of Earth’s).

Thus, the new time period $T^{\prime }$ on the Moon is:

$$T^{\prime }=2\pi \sqrt{\frac{L}{g_M}}=2\pi \sqrt{\frac{L}{g_E/6}}=2\pi \sqrt{\frac{6L}{g_E}}=\sqrt{6}\cdot T$$

Final Answer: The time period on the Moon will be $\sqrt{6}T$ (approximately 2.45 T).