Question

53. If I = 50 kg-m?, then how much torque will be
applied to stop it in 10 sec. Its initial angular speed
is 20 rad/sec. :-
(1) 100 N-m
(2) 150 N-m
(3) 200 N-m
(4) 250 N-m​

Answer 1

$\huge\underline\orange{Answer}$

$\large{\boxed{{\tau = 100 N-m}}}$

$\huge\underline\orange{Solution}$

Given :-

Torque = ?

Inertia (I) = 50 kg.m

Time (t) = 10 s

$\large{{\omega = 20rad/s}}$

Angular Acceleration :-

$\large{{\alpha = {\frac{\omega}{t}}}}$

$\large{{\alpha ={\frac{20}{10}}}}$

$\large{\boxed{{\alpha = 2rad/{s}^{2}}}}$

Torque :-

$\large{\boxed{{\tau =I{\alpha}}}}$

$\large{{\tau= 50×2}}$

$\huge\blue{\boxed{{\tau =100N-m}}}$

✨⛄⛄⛄✨

Answer 2

Answer:

$\huge\underline\purple{\sf Given:-}$

•inertia=50kg/m^2

•Time=10seconds

•ω=20rad/s

________❇

$\huge\underline\purple{\sf find :-}$

•Torque=?

_________

$\huge\underline\purple{\sf Solution:-}$

$\large\red{\boxed{\sf Angular acceleration(α)=ω/t}}$

α=20/10

α=>2rad/s^2

Now,

Torque(τ)=I x α

=>50 x 2

$\huge{\boxed{\sf Torque(τ)=100N/m}}$

$\huge\mathfrak\orange{Thanks}$