A heavy turntable, used for rotating large objects, is a solid cylindrical wheel that can rotate about its central axle with negligible fric

Question

A heavy turntable, used for rotating large objects, is a solid cylindrical wheel that can rotate about its central axle with negligible friction. The radius of the wheel is 0.330 m. A constant tangential force of 300 N applied to its edge causes the wheel to have an angular acceleration of 0.876 rad/s2.

(a)
What is the moment of inertia of the wheel (in kg · m2)?

_____ kg · m2

(b)What is the mass (in kg) of the wheel?

_________ kg

(c)The wheel starts from rest and the tangential force remains constant over a time period of 6.00 s. What is the angular speed (in rad/s) of the wheel at the end of this time period?

________ rad/s

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Ben Gia 1 year 2021-09-01T21:44:39+00:00 1 Answers 23 views 0

Answers ( )

  1. Nem
    0
    2021-09-01T21:46:15+00:00
    Reply

    Answer:

    a) [tex]I = 113.014\,kg\cdot m^{2}[/tex], b) [tex]m = 2075.556\,kg[/tex]

    Explanation:

    a) The turntable has the following physical model by using Newton’s laws:

    [tex]F \cdot R = I \cdot \alpha[/tex]

    The moment of inertia is:

    [tex]I = \frac{F\cdot R}{\alpha}[/tex]

    [tex]I = \frac{(300\,N)\cdot(0.33\,m)}{0.876\,\frac{rad}{s^{2}} }[/tex]

    [tex]I = 113.014\,kg\cdot m^{2}[/tex]

    b) The moment of inertia for a solid cylinder:

    [tex]I = \frac{1}{2}\cdot m \cdot R^{2}[/tex]

    The mass of the turntable is:

    [tex]m = \frac{2 \cdot I}{R^{2}}[/tex]

    [tex]m = \frac{(2)\cdot (113.014\,kg\cdot m^{2})}{(0.33\,m)^{2}}[/tex]

    [tex]m = 2075.556\,kg[/tex]

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