A heavy solid disk rotating freely and slowed only by friction applied at its outer edge takes 120 seconds to come to a stop. If the

Question

A heavy solid disk rotating freely and slowed only by friction applied at its outer edge takes 120 seconds to come to a stop.
If the disk had twice the radius and twice the mass, but the frictional force remained the same, the time it would it take the wheel to come to a stop from the same initial rotational speed is ____.

in progress 0
RobertKer 4 years 2021-09-05T03:02:57+00:00 1 Answers 43 views 0

Answers ( )

  1. Calantha
    0
    2021-09-05T03:04:26+00:00
    Reply

    Answer:

    The time is 16 min.

    Explanation:

    Given that,

    Time = 120 sec

    We need to calculate the moment of inertia

    Using formula of moment of inertia

    I=\dfrac{1}{2}MR^2

    If the disk had twice the radius and twice the mass

    The new moment of inertia

    I'=\dfrac{1}{2}\times2M\times(2R)^2

    I'=8I

    We know,

    The torque is

    \tau=F\times R

    We need to calculate the initial rotation acceleration

    Using formula of acceleration

    \alpha=\dfrac{\tau}{I}

    Put the value in to the formula

    \alpha=\dfrac{F\times R}{\dfrac{1}{2}MR^2}

    \alpha=\dfrac{2F}{MR}

    We need to calculate the new rotation acceleration

    Using formula of acceleration

    \alpha'=\dfrac{\tau}{I'}

    Put the value in to the formula

    \alpha=\dfrac{F\times R}{8\times\dfrac{1}{2}MR^2}

    \alpha=\dfrac{2F}{8MR}

    \alpha=\dfrac{\alpha}{8}

    Rotation speed is same.

    We need to calculate the time

    Using formula angular velocity

    \Omega=\omega'

    \alpha\time t=\alpha'\times t'

    Put the value into the formula

    \alpha\times120=\dfrac{\alpha}{8}\times t'

    t'=960\ sec

    t'=16\ min

    Hence, The time is 16 min.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )