A 5.00-kg mass on the end of a light string rotates in a circular motion on a horizontal frictionless desk. The radius of the circle is 0.80

Question

A 5.00-kg mass on the end of a light string rotates in a circular motion on a horizontal frictionless desk. The radius of the circle is 0.800 m. The string has a breaking strength of 50.0 N. What is the fastest rotational speed of the mass that will not break the string?

in progress 0
Nem 2 months 2021-08-15T05:30:48+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-08-15T05:32:21+00:00

    Answer:

    The fastest rotational speed of the mass is 3.536s^{-1}.

    Explanation:

    Here the string’s breaking strength of 50.0 N means that the centripetal force exerted on the 5.00kg mass cannot exceed 50.0N—if it does, the string would break.

    Therefore, we demand that

    (1).\;\: \dfrac{mv^2}{R} = 50.0N

    where m = 5.00kg, R = 0.800m is the radius of the circle (also the length of the string), and v is the tangential velocity of the mass.

    Now, the tangential velocity can be written in terms of the rotational speed \omega as follows:

    (2).\;\:  v = \omega R,

    and putting that into equation (1) we get:

    \dfrac{m(\omega R )^2}{R} = 50.0N

    m \omega^2 R = 50.0N,

    and we solve for the rotational speed \omega to get:

    \omega = \sqrt{\dfrac{50N}{mR} }.

    Finally, we out in the numeral values and get;

    \omega = \sqrt{\dfrac{50N}{(5.00kg)(0.800m)} }.

    \boxed{\omega = 3.536s^{-1}}

    which is the fastest rotational speed of the mass.

Leave an answer

Browse

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