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?

(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?

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#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

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

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#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

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#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=m_{b}r^{2}ω+(1/2)m_{D}r^{2}ω

0=.015(.07)r^{2}+(1/2)(.095)ωr^{2}

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

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#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

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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