
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
580.8k+ 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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

