
Average torque on a projectile of mass m, initial speed u and angle of projection θ between initial and final positions P and Q as shown in the figure about the point of projection is:
A. \[\dfrac{{m{u^2}\sin 2\theta }}{2}\]
B. \[m{u^2}\cos \theta \]
C. \[{u^2}\sin \theta \]
D. \[\dfrac{{m{u^2}\cos \theta }}{2}\]
Answer
590.4k+ views
Hint: We should know the definition of torque, angular momentum etc.
We should know the relationship between angular momentum and torque.
Complete step by step answer:
Average Torque would just be the time, average of that varying quantity averaged over the time it takes to complete a revolution. The basic equation between torque and angular momentum is, $\tau .\Delta {\text{t = }}\Delta {\text{L}}$
We know that the basic equation between torque and angular momentum is,
$\tau .\Delta {\text{t = }}\Delta {\text{L}}$
And angular momentum is the quantity of rotation of a body, which is the product of its moment of inertia and its angular velocity.
Torque is a force that caused machinery, etc. to turn \[\left( {rotate} \right)\]
We know from basic equation that,
$\tau .\Delta {\text{t = }}\Delta {\text{L}}$
$
\Delta t{\text{ = }}\dfrac{{2u\sin \theta }}{g}\left[ {u = speed,\theta = angle\;{\text{of projection}}} \right] \\
\Delta L = \left[ {{L_f} - {L_i}} \right]\;\left[ {{L_f} = find\;{\text{momentum}}} \right]\;\left[ {{L_i} = initial\;{\text{momentum}}} \right] \\
$
Initial momentum is zero as it starts form a static point thus,
$
\because \Delta L = \left( {{L_f}} \right) = m.u\sin \theta \times R \\
mv\sin \theta \times \dfrac{{{u^2}\sin 2\theta }}{g} \\
\dfrac{{m{u^3}\sin \theta .\sin 2\theta }}{g} \\
$
Therefore we know that,
$
{\tau _{ag}} = \dfrac{{\Delta L}}{{\Delta t}} \\
\dfrac{{m{u^3}\sin \theta .\sin 2\theta }}{{g.\dfrac{{2u\sin \theta }}{g}}} = \dfrac{{m{u^2}\sin 2\theta }}{2} \\
$
So, the correct answer is “Option A”.
Note:
We should take care of substitutions of different equations in one.
We should remember all basic equations.
We should make sure to avoid substitution errors.
We should know the relationship between angular momentum and torque.
Complete step by step answer:
Average Torque would just be the time, average of that varying quantity averaged over the time it takes to complete a revolution. The basic equation between torque and angular momentum is, $\tau .\Delta {\text{t = }}\Delta {\text{L}}$
We know that the basic equation between torque and angular momentum is,
$\tau .\Delta {\text{t = }}\Delta {\text{L}}$
And angular momentum is the quantity of rotation of a body, which is the product of its moment of inertia and its angular velocity.
Torque is a force that caused machinery, etc. to turn \[\left( {rotate} \right)\]
We know from basic equation that,
$\tau .\Delta {\text{t = }}\Delta {\text{L}}$
$
\Delta t{\text{ = }}\dfrac{{2u\sin \theta }}{g}\left[ {u = speed,\theta = angle\;{\text{of projection}}} \right] \\
\Delta L = \left[ {{L_f} - {L_i}} \right]\;\left[ {{L_f} = find\;{\text{momentum}}} \right]\;\left[ {{L_i} = initial\;{\text{momentum}}} \right] \\
$
Initial momentum is zero as it starts form a static point thus,
$
\because \Delta L = \left( {{L_f}} \right) = m.u\sin \theta \times R \\
mv\sin \theta \times \dfrac{{{u^2}\sin 2\theta }}{g} \\
\dfrac{{m{u^3}\sin \theta .\sin 2\theta }}{g} \\
$
Therefore we know that,
$
{\tau _{ag}} = \dfrac{{\Delta L}}{{\Delta t}} \\
\dfrac{{m{u^3}\sin \theta .\sin 2\theta }}{{g.\dfrac{{2u\sin \theta }}{g}}} = \dfrac{{m{u^2}\sin 2\theta }}{2} \\
$
So, the correct answer is “Option A”.
Note:
We should take care of substitutions of different equations in one.
We should remember all basic equations.
We should make sure to avoid substitution errors.
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