A physics student is standing on an initially motionless, frictionless turntable

Question

A physics student is standing on an initially motionless, frictionless turntable with rotational inertia 0.31 kg?m2. She's holding a wheel with rotational inertia 0.22 kg?m2spinning at 130 rpm about a vertical axis, as in the figure. (Figure 1) When she turns the wheel upside down, student and turntable begin rotating at 70 rpm.

Part A

Find the student's mass, considering her to be a 30-cm-diameter cylinder.

Express your answer using two significant figures.

nothing

SubmitRequest Answer

Part B

Neglecting the distance between the axes of the turntable and wheel, determine the work she did in turning the wheel upside down.

Express your answer using two significant figures.

A physics student is standing on an initially motionless, frictionless turntable with rotational inertia 0.31 kg?m2. She's holding a wheel with rotational inertia 0.22 kg?m2spinning at 130 rpm about a vertical axis, as in the figure. (Figure 1) When she turns the wheel upside down, student and turntable begin rotating at 70 rpm.

Part A

Find the student's mass, considering her to be a 30-cm-diameter cylinder.

Express your answer using two significant figures.

M =

nothing

  kg  

SubmitRequest Answer

Part B

Neglecting the distance between the axes of the turntable and wheel, determine the work she did in turning the wheel upside down.

Express your answer using two significant figures.

Explanation / Answer

turntable with moment of inertia :- 0.31 kg.m^2

wheel rotational inertia :- 0.22 kg.m^2

w = 130 rpm

turntable speed = 70 rpm

a)

since the turntable is frictioness there is no external torque above it's axis

Lo = 0.22 kg.m^2 x 130 rpm

Lz wheel is inverted

Lz - Lo = Lo

Lz = 2Lo

Lz = Itotal*70 rpm = 2 x 0.22 kg.m^2 x 130 rpm

Itotal = 0.817 kg.m^2

Now, Itotal included inertia of the wheel 0.31 kg.m^2 the student 1/2M(0.15 m^2) and CM of mass wheel is neglected if the axis of the wheel is turnable

Therfore,

1/2M*(0.15)^2 = (0.817 - 0.31)

M = 45.1 kg

b)

The work done by the student’s muscle equals the change in kinetic energy,

Wnc = ?K = 1/2I*w^2

Wnc = (1 / 2)(0.817 kg ? m )(70? / 30 s)2

= 22 J

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