A thin uniform rod has a length of 0.500 m and is rotating in a circle on a fric

Question

A thin uniform rod has a length of 0.500 m and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.41 rad/s and a moment of inertia about the axis of 3.20×103 kgm2 . A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is 0.146 m/s . The bug can be treated as a point mass.

Part A. What is the mass of the rod?

Part B. What is the mass of the bug?

Explanation / Answer

a) Moment of inertia of a rod about an endpoint
is
I = 1/3*m*L^2
m = 3I/L^2
m = (3*3.2*10^-3)/0.5^2
m = 0.0384 kg
Mass of rod , m = 0.0384 Kg
-------
b)
The angular momentum L is conserved:
New angular velocity is
w = v/R = 0.146 m/s /0.5 m = 0.292 rad/s
Moment of Inertia of Bug, I(bug) = mL^2

L1 = I*w1 = 3.2*10^-3*0.41
L2 = I*w2 + I(bug)*w2 = 3.2*10^-3*0.292 + m(bug)*0.5^2*0.292


L1 = L2
3.2*10^-3*0.41 = 3.2*10^-3*0.292 + m(bug)*0.5^2*0.292
m(bug) = (3.2*10^-3*0.41 - 3.2*10^-3*0.292)/(0.5^2*0.292)
m(bug) = 0.005173 kg
m(bug) = 5.173 g
Mass of bug = 5.173 g   

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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