the triceps muscle in the back of the upper arm extends the forearm. this muscle in a professional boxer exerts a force of 2.00×103 n with a

Question

the triceps muscle in the back of the upper arm extends the forearm. this muscle in a professional boxer exerts a force of 2.00×103 n with an effective perpendicular lever arm of 3.10 cm, producing an angular acceleration of the forearm of 121 rad/s2. what is the moment of inertia of the boxer’s forearm?

in progress 0
Verity 1 month 2021-08-01T22:49:00+00:00 1 Answers 17 views 0

Answers ( )

    0
    2021-08-01T22:50:59+00:00

    Answer: 0.512 kgm²

    Explanation:

    Given

    Force, F = 2*10^3 N

    Angular acceleration, α = 121 rad/s²

    Lever arm, r(⊥) = 3.1 cm = 3.1*10^-2 m

    τ = r(⊥) * F

    Also,

    τ = Iα

    Using the first equation, we have

    τ = r(⊥) * F

    τ = 0.031 * 2*10^3

    τ = 62 Nm

    Now we calculate for the inertia using the second equation

    τ = Iα, making I subject of formula, we have

    I = τ / α, on substituting, we have

    I = 62 / 121

    I = 0.512 kgm²

    Thus, the moment of inertia of the boxers forearm is 0.512 kgm²

Leave an answer

Browse

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