A 10 , kg ball attached to the end of a rigid | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: Mass $m = 10 \, kg$,length of rod $L = 1 \, m$,radius $r = 0.5 \, m$,period $T = 1.57 \, s \approx \pi/2 \, s$.First,find the angle $	heta

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(B) Given: Mass $m = 10 \, kg$,length of rod $L = 1 \, m$,radius $r = 0.5 \, m$,period $T = 1.57 \, s \approx \pi/2 \, s$.First,find the angle $	heta$ that the rod makes with the vertical. From the geometry,$\sin 	heta = r/L = 0.5/1 = 0.5$,so $	heta = 30^\circ$ or $\pi/6$ radians.The angular velocity is $\omega = 2\pi/T = 2\pi / (1.57) \approx 2\pi / (\pi/2) = 4 \, rad/s$.The forces acting on the ball are gravity ($mg$ downwards) and the force exerted by the rod $(\vec{F}_{rod})$. Let $\vec{F}_{rod}$ have components $F_v$ (vertical) and $F_h$ (horizontal).In the vertical direction: $F_v - mg = 0 \implies F_v = mg = 10 	imes 9.8 = 98 \, N$.In the horizontal direction (centripetal force): $F_h = m\omega^2 r = 10 	imes (4)^2 	imes 0.5 = 10 	imes 16 	imes 0.5 = 80 \, N$.The total force exerted by the rod is $F = \sqrt{F_v^2 + F_h^2} = \sqrt{98^2 + 80^2} = \sqrt{9604 + 6400} = \sqrt{16004} \approx 126.5 \, N$. Given the options,the intended calculation likely uses $g = 10 \, m/s^2$,giving $F_v = 100 \, N$ and $F_h = 80 \, N$,so $F = \sqrt{100^2 + 80^2} = \sqrt{16400} \approx 128 \, N$.

$A$ $10 \, kg$ ball attached to the end of a rigid massless rod of length $L = 1 \, m$ rotates at a constant speed in a horizontal circle of radius $r = 0.5 \, m$ and period $T = 1.57 \, s$ as shown in the figure. The force exerted by the rod on the ball is ........ $N$.

Similar Questions

Assertion $(A):$ Two identical trains move in opposite senses in the equatorial plane with the same speeds relative to the Earth's surface. They have equal magnitude of normal reaction.
Reason $(R):$ The trains have different centripetal accelerations due to different speeds.

Imagine a situation in which the horizontal surface of block $M_0$ is smooth and its vertical surface is rough with a coefficient of friction $\mu$. In the above problem,choose the correct value$(s)$ of $F$ for which the blocks $M$ and $m$ remain stationary with respect to $M_0$.

$A$ balloon of total mass $M$ and a fixed size starts coming down with an acceleration $f$ $(f < g)$. The fraction of the total mass of the balloon which must be dropped from it so that it starts going up with an acceleration of $f$ (assuming negligible air resistance) is

In which of the following graphs,the total change in momentum is zero?

What is the acceleration of the block and trolley system shown in the figure,if the coefficient of kinetic friction between the trolley and the surface is $0.04$? What is the tension in the string? (Take $g = 10\; m s^{-2}$). Neglect the mass of the string.

Vedclass Test Series

Exam Paper Generator

Online Exam Module