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A wheel of moment of inertia 2.5 kg m^2 ...

A wheel of moment of inertia 2.5 kg `m^2` has an initial angular velocity of 40 rad `s^(-1)`. A constant torque of 10 N m acts on the wheel. The time during which the wheel is accelerated to 60 rad `s^(-1)` is

A

4.5 s

B

6 s

C

5 s

D

2.5 s

Text Solution

Verified by Experts

The correct Answer is:
C
Here , I = `2.5 kg m^(2) , omega_(0) = 40 rad s^(-1)`
`tau = 10 N m , omega = 60 rad s^(-1)`
As `tau = I alpha or alpha = (tau)/(I) = (10)/(2.5) = 4 rad s^(-2)`
Using `omega = omega_(0) + alpha t , t = (omega - omega_(0))/(alpha)`
Substituting the values , we get `t = (60 - 40)/(4 ) = 5`s
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Knowledge Check

  • A wheel of moment of inertia 2.5 kgm^(2) has an initial angular velocity of 40 rad s^(-1) . A constant torque of 10 Nm acts on the wheel. The time during which the wheel is accelerated to 60 rad s^(-1) is:

    A
    4s
    B
    6s
    C
    5s
    D
    2.5s

  • A wheel whose moment of inertia is 2 kg m^(2) has an initial angular velocity of 50 rad/s. A constant torque of 10 Nm acts on the wheel. The time in which the wheel is accelerated to 80 rad's is

    A
    12 s
    B
    3 s
    C
    6 s
    D
    9 s

  • A flywheel of moment of inertia 3 xx 10^(2) kg m^(2) is rotating with uniform angular speed of 4.6 rad s^(-1) . If a torque of 6.9 xx 10^(2)Nm retards the wheel, then the time in which the wheel comes to rest is

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    C
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