Motion in a Plane

To solve this problem, we need to understand the relationship between the tension in the string and the speed of the bob whirling in a horizontal circle. The centripetal force acting on the bob is provided by the tension in the string, and it can be given by the formula:

$ F = \frac{m v^2}{r} $

where:

• F is the centripetal force (or tension in the string)
• m is the mass of the bob
• v is the tangential speed of the bob
• r is the radius of the circle

According to the problem, the initial speed of the bob is $10 \mathrm{~rpm}$, and the radius is kept constant. Let’s denote the initial tension in the string as $T_1$ and the new tension as $T_2$. Given that the tension in the string is quadrupled, we have:

$ T_2 = 4 T_1 $

Also, the centripetal force can be written in terms of tension:

$ T_1 = \frac{m v_1^2}{r} $

$ T_2 = \frac{m v_2^2}{r} $

By substituting $T_2 = 4 T_1$ into the second equation, we get:

$ 4 T_1 = \frac{m v_2^2}{r} $

We know from the first equation that:

$ T_1 = \frac{m v_1^2}{r} $

Substituting this into our equation for $T_2$:

$ 4 \left(\frac{m v_1^2}{r}\right) = \frac{m v_2^2}{r} $

The masses and radii cancel out, leaving us with:

$ 4 v_1^2 = v_2^2 $

Taking the square root of both sides:

$ v_2 = 2 v_1 $

The initial speed $v_1$ is given as $10 \mathrm{~rpm}$. Therefore, the new speed $v_2$ is:

$ v_2 = 2 \times 10 \mathrm{~rpm} = 20 \mathrm{~rpm} $

Hence, the new speed of the bob is 20 rpm. The correct answer is Option A.

$\begin{aligned} & \omega_1=\frac{2 \pi}{60} ; \quad x_1=\frac{2 \pi}{60} \times 60=2 \pi \\\\ & \omega_2=\frac{2 \pi}{3600} ; \quad x_2=\frac{2 \pi}{3600} \times 60=\frac{2 \pi}{60} \\\\ & \omega_3=\frac{2 \pi}{3600 \times 12} ; \quad x_3=\frac{2 \pi}{3600 \times 12} \times 60=\frac{2 \pi}{720} \\\\ & \frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=\frac{1}{60}=k \end{aligned}$

At the top most point of its trajectory particle will have only horizontal component of velocity

$\begin{aligned} & \mathrm{V} \text { at top }=v \cos \theta \\ & =20 \times \frac{1}{2} \\ & =10 \mathrm{~m} / \mathrm{s} \end{aligned}$

Direction of velocity and centripetal acceleration changes continuously so $\vec{v}$ is not constant and $\vec{a}$ is not a constant.

$\mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 g}$

$ =\frac{(280)^{2}\left(\sin 30^{\circ}\right)^{2}}{2(9.8)} $

$=1000 \mathrm{~m}$

$S=u t+\frac{1}{2} a t^{2}$

$-\mathrm{H}=4 \times 4-\frac{1}{2} \times 10 \times 4^{2}$

$-\mathrm{H}=16-80$

$-\mathrm{H}=-64$

$\mathrm{H}=64 \mathrm{~m}$

Maximum height reached by the ball

$H = {{{v^2}{{\sin }^2}\theta } \over {2g}}$

$ = {{{{(20)}^2}{{\sin }^2}30^\circ } \over {2g}}$

$H = 5\,m$

At highest point vertical component of velocity become zero.

At highest point speed of object = 10cos30$^\circ$

= 5$\sqrt3$ m/s