Consider a particle of mass m = 18.0kg revolving around an axis with angular spe

Question

Consider a particle of mass m = 18.0kg revolving around an axis with angular speed ?. The perpendicular distance from the particle to the axis is r = 1.75m . (Figure 1)

Part A: Which of the following are units for expressing rotational velocity, commonly denoted by omega?

check all that apply

radians per second

degrees per second

meters per second

arc seconds

revolutions per second

Part B: Assume omega = 17.0rad/s . What is the magnitude v of the velocity of the particle in m/s?

Part C: Now that you have found the velocity of the particle, find its kinetic energy K.

Part D: Find the moment of inertia of the particle described in the problem introduction with respect to the axis about which it is rotating. Assume omega = 17.0rad/s .

Consider a particle of mass m = 18.0kg revolving around an axis with angular speed ?. The perpendicular distance from the particle to the axis is r = 1.75m . (Figure 1) Part A: Which of the following are units for expressing rotational velocity, commonly denoted by omega? check all that apply radians per second degrees per second meters per second arc seconds revolutions per second Part B: Assume omega = 17.0rad/s . What is the magnitude v of the velocity of the particle in m/s? Part C: Now that you have found the velocity of the particle, find its kinetic energy K. Part D: Find the moment of inertia of the particle described in the problem introduction with respect to the axis about which it is rotating. Assume omega = 17.0rad/s .

Explanation / Answer

a)

1.radians per second

2.

degrees per second

3.

revolutions per second

b)

V=r*W= 17*1.75

V=29.75 m/s

c)

Moment of inertia

I=mr2=18*1.752=55.125 Kg-m2

KE=(1/2)IW2=(1/2)*55.125*172

KE=7965.6 J

d)

I=mr2=18*1.752=55.12 Kg-m2

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