Torque

An automobile engine develops $100 \mathrm{kW}$ power when rotating with an angular speed of $250 \mathrm{rad} {\mathrm{s}}^{-1}.$ The torque developed by it will be about, 

A wheel of the moment of inertia ${10}^3 \mathrm{kg}{ \mathrm{m}}^2$ is rotating with an angular speed of $5 \mathrm{rad} {\mathrm{s}}^{-1}$ The torque required to stop it in $2 \mathrm{s}$ is, 

A force $\overrightarrow{\mathrm{F}}=\left(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-5\hat{\mathrm{k}}\right) \mathrm{N}$ acts at a point ${\overrightarrow{\mathrm{r}}}_1=(2\hat{\mathrm{i}}+4\hat{\mathrm{j}}+7\hat{\mathrm{k}}) \mathrm{m}.$ The torque of the force about the point ${\overrightarrow{\mathrm{r}}}_2=(\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}) \mathrm{m}$ is

A wheel having moment of inertia $4 \mathrm{kg}{ \mathrm{m}}^2$ about its symmetrical axis, rotates at rate of $240 \mathrm{rpm}$ about it. The torque which can stop the rotation of the wheel in one minute is

Moment of a force of magnitude $10 \mathrm{N}$ acting along positive $y$-direction at point $\left(2 \mathrm{m},0,0\right)$ about the point $\left(0,1 \mathrm{m},0\right)$ in $\mathrm{N}-\mathrm{m}$ is

A cube of side $a$ is placed on an inclined plane of inclination $\theta$. What is the maximum value of $\theta$ for which the cube will not topple?

The instantaneous angular position of a point on the rotating wheel is given by the equation, $\theta \left(t\right)=\left(2t^3-6t^2\right).$ The torque on the wheel becomes zero at

A rod of weight $W$ is supported by two parallel knife edges $A$ and $B$ and is in equilibrium in a horizontal position. The knives are at a distance $d$ from each other. The centre of mass of the rod is at distance $x$ from $A$. The normal reaction on $A$ is

A force $\overrightarrow{F}=4\hat{\mathrm{i}}-5\hat{\mathrm{j}}+3\hat{\mathrm{k}}$ is acting at a point ${\overrightarrow{r}}_1=\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}.$ The torque acting about a point ${\overrightarrow{r}}_2=3\hat{\mathrm{i}}-2\hat{\mathrm{j}}-3\hat{\mathrm{k}}$ is, 

A flywheel of the moment of inertia $3\times {10}^2 \mathrm{kg} {\mathrm{m}}^2$ is rotating with a uniform angular speed of $4.6 \mathrm{rad} {\mathrm{s}}^{-1}.$ If a torque of $6.9\times {10}^2 \mathrm{N} \mathrm{m}$ retards the wheel, then the time in which the wheel comes to rest is

A force $\overrightarrow{F }=4\hat{\mathrm{i}}-10\hat{\mathrm{j}}$ acts on a body at a point having position vector $-5\hat{\mathrm{i}}-3\hat{\mathrm{j}}$ relative to origin of co-ordinates on the axis of rotation. The torque acting on the body about the origin is

A rigid body is rotating about an axis. To stop the rotation, we have to apply

The moment of inertia of a disc of radius 0.5 m about its geometric axis is 2kg-m². If a string is tied to its circumference and a force of 10 Newton is applied, if the disc executes rotatory motion, the value of its angular velocity after 2 seconds will be :-

A uniform rod $OB$ of length $1 \mathrm{m}\text{,}$ cross-sectional area $0.012 {\mathrm{m}}^2$ and relative density $2.0$ is free to rotate about $\mathrm{O}$ in vertical plane. The rod is held with a horizontal string $AB$ which can withstand a maximum tension of $45 \mathrm{N}.$ The rod and string system is kept in water as shown in figure. The maximum value of angle $\alpha$ which the rod can make with vertical without breaking the string is

A disk having a thread tightly wound on it, is being pulled upwards by the thread such that the disk starts moving upwards with an acceleration $g$. If angular acceleration is $\alpha$ and radius $R$, then the value of $R\alpha$ is

What is the torque about the origin of the force $\left(7\widehat{i}+3\widehat{j}-5\widehat{k}\right) \mathrm{N}$ acting on a particle whose position vector is $\left(\widehat{i}-\widehat{j}+\widehat{k}\right) \mathrm{m}$?

Let F be the force acting on a particle having position vector$\overrightarrow{r}$  and $\overrightarrow{\tau }$ be the torque of this force about the origin. Then