Calculate the moment of inertia (in kg·m 2 ) of a skater given the following inf

Question

Calculate the moment of inertia (in kg·m2) of a skater given the following information.

(a) The 68.0 kg skater is approximated as a cylinder that has a 0.100 m radius.

___kg·m2

(b)The skater with arms extended is approximately a cylinder that is 60.0 kg, has a 0.100 m radius, and has two 0.800 m long arms which are 4.00 kg each and extend straight out from the cylinder like rods rotated about their ends.

__kg·m2

Question 2

You have a grindstone (a disk) that is 95.0 kg, has a 0.460-m radius, and is turning at 80.0 rpm, and you press a steel axe against it with a radial force of 12.0 N.

(a) Assuming the kinetic coefficient of friction between steel and stone is 0.10, calculate the angular acceleration (in rad/s2) of the grindstone. (Indicate the direction with the sign of your answer.)

____rad/s2

(b) How many turns (in rev) will the stone make before coming to rest?

___rev  

QUESTION 3 A rope is wrapped around the rim of a large uniform solid disk of mass 225 kg and radius 2.50 m. The horizontal disk is made to rotate by pulling on the rope with a constant force of 195 N. If the disk starts from rest what is its angular speed in rev/s at the end of 2.65 s?
___rev/s

Explanation / Answer

question (1)

(a)

moment of inertia I = (1/2)*m*r^2 = (1/2)*68*0.1^2 = 0.34 kg m^2       <<<------------answer

(b)

moment of inertia of cylinder I1 = (1/2)*m*r^2 = (1/2)*60*0.1^2 = 0.3 kg m^2

moment of inertia of arms I2 = 2*(1/3)*m*l^2 = 2*(1/3)*4*0.8^2 = 1.71 kg m^2

total moment of inertia = I1 + I2 = 1.71+0.3 = 2.01 kg m^2     <<<------------answer

===============================

Q2)

(a)

net torque = r*f = I*alpha

I = moment of inertia = (1/2)*m*r^2

r = radius

f = u*F

0.46*0.1*12 = -(1/2)*95*0.46^2*alpha

angular acceleration alpha = -0.055 rad/s^2    <<<------------answer

(b)

from relation wf^2 - wi^2 = 2*alpha*theta

wi = 80 rpm = 80*2*pi/60 rad/s

wf = 0

0-(80*2*pi/60)^2 = -2*0.055*theta

theta = 638 rad = 638/(2pi) = 1002 rev    <<<------------answer

===========================

Q3)

net torque = r*F = I*alpha

I = moment of inertia = (1/2)*m*r^2

r = radius

2.5*195 = (1/2)*225*2.5^2*alpha

angular acceleration alpha = 0.693 rad/s^2

from relation wf = wi + alpha*t

wi = 0

wf = 0 + 0.693*2.65

wf = 1.84 rad/s

wf = 1.84/(2pi) = 0.3 rev/s <<<------------answer

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