A member ABC has a mass of 2.4kg and is attached to pin support at B. An 800 gram sphere D strikes the end of the member ABC with a vertical downward velocity v1 = 3.0 m/s. knowing that L=750 mm and that the coefficient of restitution between the sphere and the Plank is 0.5. Determine immediately after impact:
a) The angular velocity of member ABC.
b)The velocity of the sphere.

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bharathparasad577 bharathparasad577 · Answer 1 · 2022-12-16T14:57:19+01:00

(a)The angular velocity v(d)' is 0.936 m/s

(b)The velocity of the sphere is 3 rad/s

Let us assume that

angular velocity of member ABC is v(d)'

velocity of the sphere = ω'

Write the moment of inertia of the member ABC

I(G) = 1/12 ML²

I(G) = 1/12*(2.4)(0.75m)²

I(G) = 0.112 kgm²

After the impact

ω = ω'(ccw)

ν'(G) = (L/4)ω' (upwards)

ν'(A) = (L/4) ω' (downwards)

Take the moments about B

m(d)v(d)(L/4) + 0 = m(d)v'(d)(L/4) + I(G)ω' + Mv'(G)(L/4)

m(d)v(d)(L/4) = m(d)v'(d)(L/4) + I(G)ω' + M(L/4)²ω'

m(d)v(d)(L/4) = m(d)v'(d)(L/4) + ] I(G) + M(L/4)² ]ω'

(0.8)(3)(0.75/4) = (0.8)v'(d)(0.75/4) + [ 0.112 + (2.4)(0.75/4)²

0.45 = 0.15v(d)' + 0.1963ω' -------------------- (eq1)

From the coefficient of restitution,

v(d)' - v(A)' = -e [ v(d) - v(A) ]

v(d)' - (L/4)ω' = -(0.5) [3-0]

v(d)' - (0.75/4)ω' = - 1.5

v(d)' - 0.1875ω' = - 1.5 ---------------------- (eq 2)

From eq 1 and eq 2

v(d)' = 0.936 m/s

ω' = 3 rad/s

Therefore, the angular velocity is 0.936 m/s and the velocity of the sphere is 3 rad/s.

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