(D) The torque required to open the door remains constant.Torque $\tau = r_1 F_1 = r_2 F_2$,where $r$ is the distance from the axis of rotation and $F

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(D) The torque required to open the door remains constant.Torque $\tau = r_1 F_1 = r_2 F_2$,where $r$ is the distance from the axis of rotation and $F$ is the applied force.Given: $r_1 = 1.6 \ m$,$F_1 = 1 \ N$,$r_2 = 0.4 \ m$.Substituting the values:$1.6 \times 1 = 0.4 \times F_2$$F_2 = \frac{1.6}{0.4} = 4 \ N$.Therefore,the required force is $4 \ N$.

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

Medium

$A$ force of $1 \ N$ is applied at the free edge of a door of width $1.6 \ m$ to open it. If the door is to be opened by applying force at a point $0.4 \ m$ away from the hinges (axis of rotation),then the force required is ...... $N$.

A
$1.2$
B
$3.6$
C
$2.4$
D
$4$

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System of Particles and Rotational Motion

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Relation between Torque and Angular acceleration and it's Application

$A$ uniform rod of length $l$ and mass $m$ is pivoted at point $A$. The rod is released from a horizontal position. If the moment of inertia of the rod about point $A$ is $ml^2/3$,then its initial angular acceleration is:

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$A$ thin uniform stick of length $l$ and mass $m$ is held horizontally with its end $B$ hinged on the edge of a table. Point $A$ is suddenly released. The acceleration of the centre of mass of the stick at the time of release is:

Difficult

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If there is a change of angular momentum from $1\,J\cdot s$ to $5\,J\cdot s$ in $5\,s$,then the torque is:

MediumAIIMS 1997

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$A$ flywheel of mass $25 \,kg$ has a radius of $0.2 \,m$. It is rotating at $240 \,rpm$. What is the torque necessary to bring it to rest in $20 \,s$?

MediumTS EAMCET 2010

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The angular velocity of a body changes from $6 \ rad \ s^{-1}$ to $21 \ rad \ s^{-1}$ in a time of $1.5 \ s$. If the moment of inertia of the body is $100 \ g \ m^2$,then the rate of change of angular momentum of the body is (in $N \ m$)

MediumAP EAMCET 2025

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$A$ force of $1 \ N$ is applied at the free edge of a door of width $1.6 \ m$ to open it. If the door is to be opened by applying force at a point $0.4 \ m$ away from the hinges (axis of rotation),then the force required is ...... $N$.

Similar Questions

$A$ uniform rod of length $l$ and mass $m$ is pivoted at point $A$. The rod is released from a horizontal position. If the moment of inertia of the rod about point $A$ is $ml^2/3$,then its initial angular acceleration is:

$A$ thin uniform stick of length $l$ and mass $m$ is held horizontally with its end $B$ hinged on the edge of a table. Point $A$ is suddenly released. The acceleration of the centre of mass of the stick at the time of release is:

If there is a change of angular momentum from $1\,J\cdot s$ to $5\,J\cdot s$ in $5\,s$,then the torque is:

$A$ flywheel of mass $25 \,kg$ has a radius of $0.2 \,m$. It is rotating at $240 \,rpm$. What is the torque necessary to bring it to rest in $20 \,s$?

The angular velocity of a body changes from $6 \ rad \ s^{-1}$ to $21 \ rad \ s^{-1}$ in a time of $1.5 \ s$. If the moment of inertia of the body is $100 \ g \ m^2$,then the rate of change of angular momentum of the body is (in $N \ m$)

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