(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

$A$ wheel with a moment of inertia of $5 \times 10^{-3} \ kg \cdot m^2$ is rotating at a rate of $20 \ rev/s$. The angular deceleration required to bring it to rest in $20 \ s$ is:

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$A$ circular disc rotating with frequency $f_0 = 1.3 \, rev/sec$ comes to a stop in $30 \, seconds$. The approximate angular acceleration is ....... $rad/sec^2$.

Medium

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To maintain a rotor at a uniform angular speed of $100\, rad\, s^{-1}$,an engine needs to transmit a torque of $100\, Nm$. The power of the engine is

Easy

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$A$ 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

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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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$A$ 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

$A$ wheel with a moment of inertia of $5 \times 10^{-3} \ kg \cdot m^2$ is rotating at a rate of $20 \ rev/s$. The angular deceleration required to bring it to rest in $20 \ s$ is:

$A$ circular disc rotating with frequency $f_0 = 1.3 \, rev/sec$ comes to a stop in $30 \, seconds$. The approximate angular acceleration is ....... $rad/sec^2$.

To maintain a rotor at a uniform angular speed of $100\, rad\, s^{-1}$,an engine needs to transmit a torque of $100\, Nm$. The power of the engine is

$A$ 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 $\,N$)

$A$ 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 $20 \,s$?

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