A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can rotate with n

Question

A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can rotate with negligible friction about a stationary horizontal axis. The drum is not a uniform cylinder and has unknown moment of inertia. When you release the bucket from rest, you find that it has a downward acceleration of magnitude a. What is the tension in the cable between the drum and the bucket

in progress 0
Maris 4 years 2021-07-24T13:26:17+00:00 1 Answers 71 views 0

Answers ( )

  1. Helga
    0
    2021-07-24T13:27:52+00:00
    Reply

    Answer:

    T = mg – (m²g/(I/R² + m))

    Explanation:

    Let T be the tension in the cable between the drum and the bucket

    Now, by applying newton’s second law of gravity on the downward movement of the bucket, we will obtain;

    mg – T = ma – – – – (eq1)

    Now, on the drum , a torque of TR will be acting which will create an angular acceleration of “α” in it.

    Where R is the radius.

    Let “I” denote the moment of inertia of the drum. Thus, we have;

    TR = Iα

    Now, the angular acceleration is expressed in the form;

    α = a/R

    Where a is the linear downward acceleration.

    Thus;

    TR = Ia/ R

    T =  Ia/ R²

    Let’s put Ia/ R² for T into equation 1 to give;

    mg – Ia/R² = ma

    Ia/R² + ma = mg

    a( I/R² + m) = mg

    a = mg/(I/R² +m)

    Now putting mg/(I/R² +m) for a in eq 1 gives;

    mg – T = m(mg/(I/R² +m))

    T = mg – m(mg/(I/R² +m))

    T = mg – m²g/(I/R² + m)

Leave an answer

Browse

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