Question

which of the torques given below is required to change the angular momentum of a flywheel from 15 kgm^2/s^2 to 150 kgm^2/s^2 during a time interval of 5 seconds? a) 25 kgm^2/s^2 b) 27 kgm^2/s^2 c) 29 kgm^2/s^2 d) 31 kgm^2/s^2

Answer 1

GiveN :

• Initial momentum $\sf{L_1 \: = \: 15 \: kg m^2 s^{-1}}$
• Final Momentum $\sf{L_2 \: = \: 150 \: kg m^2 s^{-1}}$
• Time interval (t) = 5 s

To FinD :

• Torque $\tau$ required to change momentum of fly wheel.

SolutioN :

As we know that :

$\longrightarrow \boxed{\sf{\tau \: = \: \dfrac{dL}{dt}}}$

Where, dL is change in momentum. We can also write it as :

$\longrightarrow \sf{dL \: Final \: momentum \: - \: Initial momentum} \\ \\ \longrightarrow \sf{dL \: = \: L_2 \: - \: L_1}$

__________________

$\longrightarrow \sf{\tau \: = \: \dfrac{L_2 \: - \: L_1}{t}} \\ \\ \longrightarrow \sf{\tau \: = \: \dfrac{150 \: - \: 15}{5}} \\ \\ \longrightarrow \sf{\tau \: = \: \dfrac{135}{5}} \\ \\ \longrightarrow \sf{\tau \: = \: 27} \\ \\ \underline{\underline{\sf{Torque \: required \: to \: change \: momentum \: is \: 27 \: kg m^2 s^{-2}}}}$

Answer 2

Given ,

• Change in angular momentum (dL) = 135 kg•m²/s
• Time (dT) = 5 sec

As we know that ,

The rate of change of angular momentum is called torque

It is denoted by " τ " and read as a " Tau "

$\boxed{\sf{\tau \: = \: \dfrac{dL}{dt}}}$

The SI unit of torque is " N•m "

It is vector quantity

Thus ,

τ = 135/5

τ = 27 N•m

$\sf \therefore{ \underline{The \: required \: torque \: is \: 27 \: Nm}}$