Suppose a 0.250 kg ball is thrown at 16.0 m/s to a motionless person standing on

Question

Suppose a 0.250 kg ball is thrown at 16.0 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in Figure 9.31.

(a) Calculate the final linear velocity of the person, given his mass is 72.0 kg.

(b) What is his angular velocity if each arm has a 5.00 kg mass? You may treat his arms as uniform rods of length 0.9 m and the rest of his body as a uniform cylinder of radius 0.170 m. Neglect the effect of the ball on his rotational inertia and on his center of mass, so that it remains in his geometrical center.

(c) Compare the initial and final total kinetic energy.

Explanation / Answer

applying the law of consservation of momentum

mu = (m+M) V

0.25 * 16 = (0.25+72)*V

V = .0.0553 m/s

b) apply the law of conservation of angula rmomentum

as L1 = L2

mvr = (I1 + 2I2 + I3)w

I1 = body MI = MR^2/2 = (72)*0.17^2/2 = 1.0404

I2 = arm MI = mL^2/3 = 5*.0.9^2/3 = 1.35

I3 = 0.25 * 0.9^2 = .2025

0.25*16*0.9 = (1.0404 + 1.35 + 0.2025) W

W = 1.388 rad/s
------------------------------------------------------------------
c)

Initial Kinetic energy = 0.5*0.25*16^2 = 32 J

Final KE = translational KE + rotational KE

KE f = 0.5*(72+0.25)*0.0553^2 + 0.5(1.0404 + 0.2025 + 1.35) *1.388^2 = 2.608

Initial / final = 32/2.608 = 12.27

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.