a) Define angular momentum. Write its SI unit. [1+1] b) The speed of a motor engine decreases from 900\ rev/min. to 600rev/min. in 10\ seconds . Calculate : [3] i) The angular acceleration ii)…

a) Define angular momentum. Write its SI unit.

Angular momentum is the rotational analog of linear momentum. It is defined as the product of the moment of inertia ($I$) of a rotating body and its angular velocity ($\omega$). Mathematically, it is expressed as: $$L=I\omega$$ where:

• $L$ = angular momentum,
• $I$ = moment of inertia,
• $\omega$ = angular velocity.

SI unit of angular momentum is kilogram meter squared per second (kg·m²/s).

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b) The speed of a motor engine decreases from $900rev/min.$ to $600rev/min.$ in $10\mathrm{seconds}$.

i) The angular acceleration

Given:

• Initial angular velocity, ${\omega }_0=900rev/min$
• Final angular velocity, $\omega =600rev/min$
• Time, $t=10\mathrm{s}$

Step 1: Convert angular velocities from revolutions per minute (rev/min) to radians per second (rad/s). $$1\mathrm{rev}=2\pi \mathrm{rad}$$ $${\omega }_0=900\times \frac{2\pi }{60}=30\pi rad/s$$ $$\omega =600\times \frac{2\pi }{60}=20\pi rad/s$$

Step 2: Use the formula for angular acceleration ($\alpha$): $$\alpha =\frac{\omega -{\omega }_0}{t}$$ $$\alpha =\frac{20\pi -30\pi }{10}=\frac{-10\pi }{10}=-\pi rad/{\mathrm{s}}^2$$

Angular acceleration = $-3.14rad/{\mathrm{s}}^2$ (taking $\pi \approx 3.14$).

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ii) Number of revolutions made by the motor during this interval

Step 1: Use the formula for angular displacement ($\theta$): $$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$$ $$\theta =(30\pi )(10)+\frac{1}{2}(-\pi )(10{)}^2$$ $$\theta =300\pi -50\pi =250\pi \mathrm{rad}$$

Step 2: Convert angular displacement to revolutions: $$\text{Number of revolutions}=\frac{\theta }{2\pi }=\frac{250\pi }{2\pi }=125$$

Number of revolutions = 125.

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iii) How many additional seconds are required for motor to come to rest in the same rate

Given:

• Final angular velocity ($\omega$) = 0 rad/s
• Initial angular velocity (${\omega }_0$) = 20π rad/s (since the motor is already slowing down to 600 rev/min)
• Angular acceleration ($\alpha$) = -π rad/s² (same rate)

Step 1: Use the formula for time ($t$): $$\omega ={\omega }_0+\alpha t$$ $$0=20\pi +(-\pi )t$$ $$\pi t=20\pi$$ $$t=20\mathrm{s}$$

Additional seconds required = 20.

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OR

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a) For a simple pendulum, show that the acceleration is directly proportional to the displacement of bob from its mean position.

Step 1: Consider a simple pendulum of length $L$ and mass $m$ displaced by a small angle $\theta$ from its mean position.

Step 2: The restoring force ($F$) acting on the bob is given by: $$F=-mg\sin \theta$$ For small angles, $\sin \theta \approx \theta$ (in radians), so: $$F\approx -mg\theta$$

Step 3: The acceleration ($a$) of the bob is given by Newton’s second law: $$a=\frac{F}{m}=\frac{-mg\theta }{m}=-g\theta$$

Step 4: The displacement ($x$) of the bob from its mean position is: $$x=L\theta$$ $$\theta =\frac{x}{L}$$

Step 5: Substitute $\theta$ in the acceleration equation: $$a=-g\left(\frac{x}{L}\right)=-\frac{g}{L}x$$

Conclusion: The acceleration ($a$) is directly proportional to the displacement ($x$) of the bob from its mean position, with the proportionality constant being $-\frac{g}{L}$.

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b) A bob of mass 8 kg performs SHM of amplitude 30 cm. The restoring force is 60 N.

i) Time period

Given:

• Mass ($m$) = 8 kg
• Amplitude ($A$) = 30 cm = 0.3 m
• Restoring force ($F$) = 60 N

Step 1: Use Hooke’s law for SHM: $$F=-kx$$ where $k$ is the force constant.

Step 2: At maximum displacement ($x=A$): $$60=k(0.3)$$ $$k=\frac{60}{0.3}=200N/m$$

Step 3: The time period ($T$) of SHM is given by: $$T=2\pi \sqrt{\frac{m}{k}}$$ $$T=2\pi \sqrt{\frac{8}{200}}=2\pi \sqrt{\frac{1}{25}}=2\pi \times \frac{1}{5}=\frac{2\pi }{5}$$

Time period = $1.26\mathrm{s}$ (taking $\pi \approx 3.14$).

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ii) The maximum acceleration

Step 1: The maximum acceleration ($a_{\text{max}}$) occurs at maximum displacement ($x=A$): $$a_{\text{max}}={\omega }^{2}A$$ where $\omega$ is the angular frequency.

Step 2: Calculate $\omega$: $$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{8}}=\sqrt{25}=5rad/s$$

Step 3: Calculate $a_{\text{max}}$: $$a_{\text{max}}=(5{)}^2(0.3)=25\times 0.3=7.5m/{\mathrm{s}}^2$$

Maximum acceleration = $7.5m/{\mathrm{s}}^2$.

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iii) Kinetic energy when displacement is 12 cm

Given:

• Displacement ($x$) = 12 cm = 0.12 m

Step 1: The total energy ($E$) of SHM is conserved and given by: $$E=\frac{1}{2}k{A}^2$$ $$E=\frac{1}{2}\times 200\times (0.3{)}^2=100\times 0.09=9\mathrm{J}$$

Step 2: The potential energy ($PE$) at displacement $x$ is: $$PE=\frac{1}{2}k{x}^2=\frac{1}{2}\times 200\times (0.12{)}^2=100\times 0.0144=1.44\mathrm{J}$$

Step 3: The kinetic energy ($KE$) is: $$KE=E-PE=9-1.44=7.56\mathrm{J}$$

Kinetic energy = $7.56\mathrm{J}$.