Inte ngure,·8.47 g bulet is fired ineo . 0.516kg block attached to the end of a

Question

Inte ngure,·8.47 g bulet is fired ineo . 0.516kg block attached to the end of a 0.578 m nonuniform rod of mass 0.406 kg. The block-rod-bullet system then rotates in the plane of the figure, about a fixed axis at A. The rotational inertis of the rod alone about A is 0.0764 kg-m2. Treat the block as a particle. (a) What then is the rotational inertia of the Just before impact? system about point Arb) ?f the angular speed of the system about A just after impact is 6. 18 rad/s, what is the bullet's speed Units

Explanation / Answer

Given that,

mass of bullet , m = 8.47 g = 0.00847 kg

mass of block , M = 0.516 kg

Length of rod, L = 0.578 m

Rotational inertia, Io = 0.0764 kg.m^2

angular speed, w = 6.18 rad/s

(a)

rotational inertia of the block-rod-bullet system about point A,

l = lo + (m + M)*L^2

l = 0.0764 + (0.00847 + 0.516)*0.578^2

l = 0.2516 kg.m^2

(b)

Let speed of block = v

From  law of conservation of angular momentum,

Li = Lf

m*v*L = l*w

0.00847 * v * 0.578 = 0.2516 * 6.18

v = 317.6 m/s

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