(D) The torque $\tau$ is given by $\tau = F r \sin \theta = I \alpha$.Here,the moment of inertia $I$ of the four masses is $I = 4 \times (m r^2) = 4 \

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

(D) The torque $\tau$ is given by $\tau = F r \sin \theta = I \alpha$.Here,the moment of inertia $I$ of the four masses is $I = 4 \times (m r^2) = 4 \times (2 \times (1/4)^2) = 4 \times (2 \times 1/16) = 0.5 \ kg \cdot m^2$.The torque applied by force $F = 24 \ N$ at a distance $r = 1/2 \ m$ with angle $\theta = 30^{\circ}$ is $\tau = F r \sin 30^{\circ} = 24 \times (1/2) \times (1/2) = 6 \ N \cdot m$.Using $\tau = I \alpha$,we get $6 = 0.5 \times \alpha$.Therefore,$\alpha = 6 / 0.5 = 12 \ rad/s^2$.

Class 11 Physics System of Particles and Rotational Motion Relation between Torque and Angular acceleration and it's Application

Difficult

Four masses of $2 \ kg$ each are attached to an axle by spokes of length $1/4 \ m$ as shown in the figure. $A$ force $F = 24 \ N$ is applied at the end of a spoke of length $1/2 \ m$ at an angle of $30^{\circ}$ to produce an angular acceleration $\alpha$. The value of $\alpha$ is ........ $rad/s^2$.

A
$4$
B
$10$
C
$20$
D
$12$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

System of Particles and Rotational Motion

Subtopic

Relation between Torque and Angular acceleration and it's Application

$A$ rod $PQ$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kept horizontal by a massless string tied to point $Q$ as shown in the figure. When the string is cut,the initial angular acceleration of the rod is

Difficult

View Solution

Write Newton's second law for a system of particles performing rotational motion.

Medium

View Solution

$A$ disc of radius $0.4 \,m$ and mass $1 \,kg$ rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration of the disc is $10 \,rad/s^2$. The tangential force applied to the rim of the disc is (in $\,N$)

MediumMHT CET 2021

View Solution

$A$ thin uniform hemispherical bowl of mass $m$ and radius $R$ is lying on a smooth horizontal surface. $A$ horizontal force $F$ is now applied perpendicular to the rim of the bowl (see figure). The instantaneous angular acceleration of the bowl will be

Difficult

View Solution

$A$ pulley has the shape of a uniform solid disc of mass $2 \ kg$ and radius $0.5 \ m$. $A$ string is wrapped over its rim and is pulled by a force of $2.5 \ N$. The pulley is free to rotate about its axis. Initially,the pulley is at rest. Find the angular velocity of the pulley just after $10 \ s$. (in $rad/s$)

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$A$ rod $PQ$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kept horizontal by a massless string tied to point $Q$ as shown in the figure. When the string is cut,the initial angular acceleration of the rod is

Write Newton's second law for a system of particles performing rotational motion.

$A$ disc of radius $0.4 \,m$ and mass $1 \,kg$ rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration of the disc is $10 \,rad/s^2$. The tangential force applied to the rim of the disc is (in $\,N$)

$A$ thin uniform hemispherical bowl of mass $m$ and radius $R$ is lying on a smooth horizontal surface. $A$ horizontal force $F$ is now applied perpendicular to the rim of the bowl (see figure). The instantaneous angular acceleration of the bowl will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module