What linear speed must a 5.5 times 10-2kg hula hoop have if its total kinetic en

Question

What linear speed must a 5.5 times 10-2kg hula hoop have if its total kinetic energy is to be 0.10J? Assume the hoop rolls on the ground without slipping. Express your answer using two significant figures. A centrifuge with an angular speed of 6200rpm produces a maximum centripetal acceleration equal to 6880(7 (that is> 6880 times the acceleration of gravity). What is the diameter of this centrifuge? What force must the bottom of the sample holder exert on a 15.0-g sample under these conditions?

Explanation / Answer

1)

A)

Total KE = 1/2 mv2 + 1/2 I?2

I = mr2 , ? = v/r

SO,

Total KE = mv2 = 0.10

m = 0.055

i.e v = 1.348 m/sec

Therefore, the answer is 1.35 m/sec

2)

A)

6880*9.8 = r *(6200*2*pi/60)^2

67424 = r*421541.77

r = 0.16 m

diameter = 2*r = 2*0.16 = 0.32 m

B)

F = m*a = 15*10^-3*6880*9.8 = 1011.36 N = 1.011 kN

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