Circular Motion

Tension in the string $=$ Centripetal force

$ \begin{aligned} \Rightarrow \quad T & =m r \omega^2=m r(2 \pi f)^2 \\ & =4 \pi^2 m r f^2 \\ & =4 \times(314)^2 \times 0.5 \times 2 \times\left(\frac{40}{60}\right)^2 \\ & =17.53 \mathrm{~N} \approx 17.5 \mathrm{~N} \end{aligned} $

$ \begin{aligned} &\text { Tension in the wire }\\ &\begin{aligned} & =\text { centripetal force }=m r \omega^2 \\ & =4 \times 2.5 \times(2 \pi f)^2=10 \times 4 \pi^2 f^2 \\ & =10 \times 4 \pi^2 \times\left(\frac{2}{\pi}\right)^2=160 \mathrm{~N} \end{aligned} \end{aligned} $

Since speed is constant throughout its entire motion on semicircular path, thus its speed is constant. Thus, change in kinetic energy is zero.

Step 1: Angular velocity of the hour hand

The hour hand of a clock completes one full revolution (360° or $2\pi$ radians) in 12 hours.

$ \omega_{\text{hour hand}} = \frac{2\pi}{12\ \text{hours}} = \frac{\pi}{6\ \text{hours}} $

Step 2: Angular velocity of Earth’s rotation

The Earth completes one full rotation (360° or $2\pi$ radians) in 24 hours.

$ \omega_{\text{Earth}} = \frac{2\pi}{24\ \text{hours}} = \frac{\pi}{12\ \text{hours}} $

Step 3: Ratio of angular velocities

$ \frac{\omega_{\text{hour hand}}}{\omega_{\text{Earth}}} = \frac{\frac{\pi}{6}}{\frac{\pi}{12}} = \frac{12}{6} = 2 $

$ \boxed{\text{Ratio } = 2 : 1} $

✅ Correct Option: B) $2 : 1$

Given:

Centripetal acceleration, $ a_c = 18 \, \mathrm{m/s}^2 $

Radius of the circular path, $ r = 50 \, \mathrm{cm} = 50 \times 10^{-2} \, \mathrm{m} $

We know the formula for centripetal acceleration:

$ a_c = \frac{v^2}{r} $

To find the velocity $ v $, rearrange the formula:

$ v = \sqrt{a_c \cdot r} $

Substituting the given values:

$ v = \sqrt{18 \times 50 \times 10^{-2}} $

$ v = \sqrt{900 \times 10^{-2}} $

$ v = 3 \, \mathrm{m/s} $

The angular velocity $ \omega $ is:

$ \omega = \frac{v}{r} = \frac{3}{50 \times 10^{-2}} $

$ \omega = 6 \, \mathrm{rad/s} $

To find the angular displacement $ \theta $ over the time interval $\frac{\pi}{18} \, \mathrm{s}$:

$ \theta = \omega t $

$ \theta = 6 \times \frac{\pi}{18} = \frac{\pi}{3} \, \mathrm{rad} $

The change in velocity due to a change in angle $ \theta $ is given by:

$ |\Delta v| = 2v \sin\left(\frac{\theta}{2}\right) $

Substituting the known values:

$ |\Delta v| = 2 \times 3 \times \sin\left(\frac{\pi}{6}\right) $

$ |\Delta v| = 2 \times 3 \times \frac{1}{2} $

$ |\Delta v| = 3 \, \mathrm{m/s} $

To find the average angular velocity of a particle moving around a circular path, we use the formula:

$ \text{Average angular velocity} = \frac{\text{Total angular displacement}}{\text{Total time}} $

For the first half of the path, which is half the circumference, the angular displacement is calculated as follows:

$ \theta_1 = \frac{\pi r}{r} = \pi \ \text{rad} $

Similarly, for the second half of the path, the angular displacement is:

$ \theta_2 = \frac{\pi r}{r} = \pi \ \text{rad} $

Thus, the total angular displacement for the entire circular path is:

$ \text{Total angular displacement} = \theta_1 + \theta_2 = 2\pi \ \text{rad} $

Given that it takes 4 seconds to travel the first half and 2 seconds for the second half, the total time is:

$ \text{Total time} = 4 + 2 = 6 \ \text{s} $

Therefore, the average angular velocity is:

$ \text{Angular velocity} = \frac{2\pi}{6} \ \text{rad/s} = \frac{\pi}{3} \ \text{rad/s} $

Drawing the diagram of the condition that is mentioned in question.

The condition is that disc is moving not sliding on hemisphere.

Here centripetal force by moving disc is balancing the weight of disc.

$ \begin{aligned} \frac{m v^2}{R} & =m g \\ v & =\sqrt{g R} \end{aligned} $

Explanation

We need to determine the number of revolutions a wheel makes before coming to a stop. Here's the breakdown of the solution:

Given Values:

Initial angular velocity, $\omega_0 = 600 \, \text{rev/min}$

To convert to radians per second:

$ \omega_0 = 600 \times \frac{2\pi}{60} = 20\pi \, \text{rad/s} $

Angular deceleration, $\alpha = -2 \, \text{rad/s}^2$

Final angular velocity, $\omega = 0 \, \text{rad/s}$

Find:

Number of revolutions, $N$

Solution:

Using the angular kinematic equation:

$ \omega^2 = \omega_0^2 + 2\alpha\theta $

Plugging in the given values:

$ 0 = (20\pi)^2 + 2(-2)\theta $

Simplify:

$ 0 = 400\pi^2 - 4\theta $

Rearrange to solve for $\theta$:

$ 4\theta = 400\pi^2 \\ \theta = 100\pi^2 $

Relate $\theta$ to revolutions:

Since $\theta = 2\pi N$,

$ 2\pi N = 100\pi^2 $

Solve for $N$:

$ N = \frac{100\pi^2}{2\pi} \\ N = 50\pi $

Assuming $\pi \approx 3.14$,

$ N = 50 \times 3.14 = 157 $

So, the number of revolutions the wheel makes before coming to rest is 157.

Given,

Mass of body, $m=10 \mathrm{~kg}$

Length of wire, $l=0.3 \mathrm{~m}$

Breaking stress, $\sigma=4.8 \times 10^7 \mathrm{~N} / \mathrm{m}^2$

Area of cross-section, $A=10^{-6} \mathrm{~m}^2$

Let angular velocity be $\omega$

$\begin{aligned} & \because \text { Stress, } \sigma=\frac{\text { Force }(F)}{A} \\ & \Rightarrow \quad F=\sigma A=m \omega^2 l \Rightarrow \omega=\sqrt{\frac{\sigma A}{m l}} \\ & \Rightarrow \quad \omega=\sqrt{\frac{4.8 \times 10^7 \times 10^{-6}}{10 \times 0.3}}=\sqrt{16}=4 \mathrm{~rad} \mathrm{~s}^{-1} \end{aligned}$

Given, mass of car, $m=500 \mathrm{~kg}$

Radius of track, $r=50 \mathrm{~m}$

Velocity of car, $v=36 \mathrm{~km} / \mathrm{h}=\frac{36 \times 10^3}{60 \times 60}=10 \mathrm{~ms}^{-1}$

Let centripetal force be $F$.

As we know that,

$\begin{aligned} & F=\frac{m v^2}{r} \\ & \therefore \quad F=\frac{500 \times 10^2}{50}=10 \times 10^2=1000 \mathrm{~N} \\ & \end{aligned}$

Given, radius of wall, R = 8 m

Minimum speed, $v=5\sqrt5$ ms$^{-1}$

Acceleration due to gravity, g = 10 ms$^{-1}$

Let coefficient of friction be $\mu$.

According to diagram,

$f=mg$ ..... (i)

$\therefore \quad f=\mu N$ ..... (ii)

where, $f$ is friction force and N is normal reaction = centripetal force

$\Rightarrow N=\frac{mv^2}{R}$

Substituting in Eq. (ii), we get

$f=\mu \frac{mv^2}{R}$ .... (iii)

Now, on equating Eqs. (i) and (iii), we get

$\begin{aligned} f & =m g=\mu \frac{m v^2}{R} \\ \Rightarrow \quad \mu & =\frac{R g}{v^2}=\frac{8 \times 10}{(5 \sqrt{5})^2}=\frac{80}{125}=0.64\end{aligned}$

As we know that,

Earth rotates from West to East and, in assertion two trains are moving in opposite direction.

$\therefore$ The train going towards East will have high speed with respect to earth from the train going towards West.

$\because$ Normal reaction is proportional to (speed)$^2$ and speeds of different trains are different.

Hence, magnitude of normal reaction is different.

$\therefore$ Assertions will be false and reason will be true because their relative speeds are different. But actual speeds of both the trains are same.