(B) The torque $\tau$ acting on the rod about the hinge is given by $\tau = F \times r$,where $r = 5L/6$ is the distance from the hinge.Given $F = Mg/

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

(B) The torque $\tau$ acting on the rod about the hinge is given by $\tau = F \times r$,where $r = 5L/6$ is the distance from the hinge.Given $F = Mg/2$,the torque is $\tau = (Mg/2) \times (5L/6) = 5MgL/12$.The moment of inertia $I$ of a rod of mass $M$ and length $L$ about an axis passing through one of its ends is $I = ML^2/3$.Using the rotational analog of Newton's second law,$\tau = I\alpha$,where $\alpha$ is the angular acceleration:$5MgL/12 = (ML^2/3) \times \alpha$$\alpha = (5MgL/12) \times (3/ML^2)$$\alpha = 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. One of its ends is hinged on a vertical axis. $A$ horizontal force $F = Mg/2$ is applied at a distance of $5L/6$ from the hinged end. What will be the angular acceleration of the rod?

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

$A$ wheel having a moment of inertia of $2 \, kg \cdot m^2$ about its vertical axis rotates at the rate of $60 \, rpm$ about the axis. The torque required to stop the wheel's rotation in one minute would be

Difficult

View Solution

$A$ constant torque of $31.4 \, Nm$ is exerted on a pivoted wheel. If the angular acceleration of the wheel is $4\pi \, rad/s^2$,then the moment of inertia will be ....... $kg \cdot m^2$.

EasyAIIMS 2001

View Solution

$A$ wheel having a moment of inertia $2 \; kg \cdot m^2$ about its vertical axis rotates at the rate of $60 \; rpm$ about this axis. The torque required to stop the wheel's rotation in one minute is:

MediumAIPMT 2004

View Solution

$A$ 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

DifficultAIPMT 2013

View Solution

The angular momentum of a wheel changes from $2L$ to $5L$ in $3$ seconds. The magnitude of the torque acting on the wheel is:

Easy

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. One of its ends is hinged on a vertical axis. $A$ horizontal force $F = Mg/2$ is applied at a distance of $5L/6$ from the hinged end. What will be the angular acceleration of the rod?

Similar Questions

$A$ wheel having a moment of inertia of $2 \, kg \cdot m^2$ about its vertical axis rotates at the rate of $60 \, rpm$ about the axis. The torque required to stop the wheel's rotation in one minute would be

$A$ constant torque of $31.4 \, Nm$ is exerted on a pivoted wheel. If the angular acceleration of the wheel is $4\pi \, rad/s^2$,then the moment of inertia will be ....... $kg \cdot m^2$.

$A$ wheel having a moment of inertia $2 \; kg \cdot m^2$ about its vertical axis rotates at the rate of $60 \; rpm$ about this axis. The torque required to stop the wheel's rotation in one minute is:

$A$ 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

The angular momentum of a wheel changes from $2L$ to $5L$ in $3$ seconds. The magnitude of the torque acting on the wheel is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module