Question

Bettle Turntable Angular Momentum and Angular Velocity Problem

A beetle with a mass of 15.0 g is initially at rest on the outer edge of a horizontal turntable that is also initially at rest. The turntable, which is free to rotatewith no friction about an axis through its center, has a mass of 95.0 g and can be treated as a uniform disk. The beetle then starts to walk around the edge of theturntable, traveling at an angular velocity of 0.0700 rad/s clockwise with respect to the turntable.

(a) With respect to you, motionless as you watch the beetle and turntable, what is the angular velocity of the beetle? Use a positive sign if the answer is clockwise,and a negative sign if the answer is counter-clockwise.
? rad/s

(b) What is the angular velocity of the turntable (with respect to you)? Use a positive sign if the answer is clockwise, and a negative sign if the answer iscounter-clockwise.
? rad/s

(c) If a mark is placed on the turntable at the beetle's starting point, how long does it take the beetle to reach the mark again?
0 0
ReportAnswer #1
Crap.. Sorry my calculation mistake.

angular velocity is actually

0.0221052632rad/s

So a) should give 0.07 - 0.0221052632=0.04789rad/s (not 0.0579... sorry)

b) should be - 0.02211rad/s
ReportAnswer #2

a) Using the conservation of angular momentum, we can solve this problem.

Let m be the mass of the beetle and M be the mass of the turntable.

mR^2ω_b + Iω_t=0, where I is the moment of inertia for the turntable, which is 1/2MR^2.

ω_t = - 2m/Mω_b = -0.07 * 2 * 15/ 95=-0.02210rad/s.

In our eyes, beetle is on the turntale. Thereforeω = 0.0700 - 0.02210 = + 0.0579rad/s

b) As we have seen in part a, it is -0.02210rad/s

c) The answer depends on the angular velocity of the beetle RELATIVE TO THE TURNTABLE. Hence

T=2π/ω= 2π/0.0700=89.76s

ReportAnswer #3

a+b) angular momentum has to be conserved. Since there is no motion at first, the later motions have to sum to 0.
0=L
0=mbr2ω+(1/2)mDr2ω

0=.015(.07)r2+(1/2)(.095)ωr2

ω=-.015(.07)/(1/2)(.095)

ω=-.022105 rad/s turn table

ω=.07-.022105

ω=.04789 rad/s of beetle respect to you

c) we have to use the angular speed of the beetle with respect to the turn table.

θ=ωt

t=θ/ω=2π/.07

t=89.76 sec

Hope that helps

ReportAnswer #4
a+b) angular momentum has to be conserved. Since there is no motion at first, the later motions have to sum to 0.
0=L
0=mbr2?+(1/2)mDr2?
0=.015(.07)r2+(1/2)(.095)?r2
?=-.015(.07)/(1/2)(.095)
?=-.022105 rad/s turn table
?=.07-.022105
?=.04789 rad/s of beetle respect to you

c) we have to use the angular speed of the beetle with respect to the turn table.
?=?t
t=?/?=2p/.07
t=89.76 sec

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