(B) The torque $\tau$ about the hinged end is given by $\tau = F \times r$,where $r = \frac{5L}{6}$.$\tau = \left( \frac{Mg}{2} \right) \times \left(

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

(B) The torque $\tau$ about the hinged end is given by $\tau = F \times r$,where $r = \frac{5L}{6}$.$\tau = \left( \frac{Mg}{2} \right) \times \left( \frac{5L}{6} \right) = \frac{5MgL}{12}$.The moment of inertia $I$ of a rod about an axis passing through one end is $I = \frac{ML^2}{3}$.Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:$\alpha = \frac{\tau}{I} = \frac{5MgL}{12} \times \frac{3}{ML^2}$.$\alpha = \frac{5g}{4L}$.

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

Difficult

$A$ rod of mass $M$ and length $L$ is placed in a horizontal plane with one end hinged about a vertical axis. $A$ horizontal force of $F = \frac{Mg}{2}$ is applied at a distance $\frac{5L}{6}$ from the hinged end. The angular acceleration of the rod will be:

A
$\frac{4g}{5L}$
B
$\frac{5g}{4L}$
C
$\frac{3g}{4L}$
D
$\frac{4g}{3L}$

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

$A$ constant torque of $1000 \; Nm$ turns a wheel of moment of inertia $200 \; kg \cdot m^2$ about an axis through its centre. The wheel is at rest initially. Its angular velocity after $3 \; s$ is

Easy

View Solution

$A$ uniform thin rod of length $L$ and mass $m$ is lying on a smooth horizontal table. $A$ horizontal impulse $P$ is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is

MediumWBJEE 2021

View Solution

$A$ flywheel of moment of inertia $2 \, kg \cdot m^2$ is rotated at a speed of $30 \, rad/s$. $A$ tangential force at the rim stops the wheel in $15 \, s$. The average torque of the force is ........... $N \cdot m$.

Medium

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

An electric motor is started from rest to produce a torque. It acquires an angular acceleration given by $\alpha = 3t - t^2$. The angular acceleration becomes zero after $2 \ s$. Calculate the angular velocity after $6 \ s$.

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

$A$ rod of mass $M$ and length $L$ is placed in a horizontal plane with one end hinged about a vertical axis. $A$ horizontal force of $F = \frac{Mg}{2}$ is applied at a distance $\frac{5L}{6}$ from the hinged end. The angular acceleration of the rod will be:

Similar Questions

$A$ constant torque of $1000 \; Nm$ turns a wheel of moment of inertia $200 \; kg \cdot m^2$ about an axis through its centre. The wheel is at rest initially. Its angular velocity after $3 \; s$ is

$A$ uniform thin rod of length $L$ and mass $m$ is lying on a smooth horizontal table. $A$ horizontal impulse $P$ is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is

$A$ flywheel of moment of inertia $2 \, kg \cdot m^2$ is rotated at a speed of $30 \, rad/s$. $A$ tangential force at the rim stops the wheel in $15 \, s$. The average torque of the force is ........... $N \cdot m$.

An electric motor is started from rest to produce a torque. It acquires an angular acceleration given by $\alpha = 3t - t^2$. The angular acceleration becomes zero after $2 \ s$. Calculate the angular velocity after $6 \ s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module