Question

A wheel starting with angular velocity of 10 radian/sec acquires angular velocity of 100
radian/sec in 15 seconds. If moment of inertia is 10kg-m2, then applied torque (in newtor
meter) is
(a) 900 (b) 100
2 1 100 25 too 10
(c) 90 (d) 60
da_13 Alandatoru​

Answer 1

Explanation:

torque =to moment of inertia into ang acceleration and angular acceleration equal to change in angular velocity divide by time

Answer 2

The applied torque in the wheel is 60 N-m.

Explanation:

Given that,

Initial angular velocity of the wheel, $\omega_i=10\ rad/s$

Final angular velocity of the wheel, $\omega_f=100\ rad/s$

Time, t = 15 s

The moment of inertia, $I=10\ kg-m^2$

We need to find the applied torque in the wheel. The torque in the rotational kinematics is given by :

$\tau=I\omega\\\\\tau=I\times (\dfrac{\omega_f-\omega_i}{t})\\\\\tau=10\times (\dfrac{100-10}{15})\\\\\tau=60\ N-m$

So, the applied torque in the wheel is 60 N-m. hence, this is the required solution.

Learn more,

Rotational kinematics

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