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A wheel is attached to a fixed shaft, and the system is free torotate without friction. To measure the moment of inertia o thewheel-shaft system, a tape of negligible mass wrapped around theshaft is pulled with a known constant force F. When a length L oftape has unwound, the system is rotating with angular speedomega-naught. Find the momentum of the inertia of the system,I-naught.

Ans. clue. If F= 10N, L= 5m, omega-naught= .5 rad/sec, thenI-naught= 400kgm^2

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

a) At the bottom is a 3 step pulley; consisting ofthree wheelsof different diameters.b) The rotating arm has two massesattached,and is screwed into the 3 step pulley. Whenthe pulleyrotates, so does the rod.c) The string is wrapped around one ofthewheels in the 3 step pulley and passes overthe smart pulley. Themass hanger is hooked into the loop at the end of the string.d)When the system is released, the mass hanger falls, turning thesmart pulley, and the 3 step pulley withthe rod attached. e)Assume: that the smart pulley is very light, so its moment ofinertia is negligible as compared to the moment of inertia of the 3step pulley and rod.



Suppose that you measured the moment of inertia of the system as I= 5.05 kg-m2. Assume the diameter of the pulley around which youwrap the string is d = 2.5 cm.

a) Suppose the system is initially spinning at angular speed ωo =9.27 rad/s. After you drop a disk of mass 856 g on it, the systemrotates at ωf = 8.35 rad/s. Find:
- the moment of inertia of the disk: kg-m2
- the radius of the disk: cm

b) suppose the system is initially spinning at angular speed ωo =2.05 rad/s. After you drop a cube of mass 786 g on it, the systemrotates at ωf = 1.79 rad/s. Find:
- the moment of inertia of the cube: kg-m2
- the length of one side of the cube: cm

a) At the bottom is a 3 step pulley; consisting ofthree wheelsof different diameters.b) The rotating arm has two massesattached,and is screwed into the 3 step pulley. Whenthe pulleyrotates, so does the rod.c) The string is wrapped around one ofthewheels in the 3 step pulley and passes overthe smart pulley. Themass hanger is hooked into the loop at the end of the string.d)When the system is released, the mass hanger falls, turning thesmart pulley, and the 3 step pulley withthe rod attached. e)Assume: that the smart pulley is very light, so its moment ofinertia is negligible as compared to the moment of inertia of the 3step pulley and rod.

a)when you drop the mass holder, the system rotates at angularacceleration α. If you wrapped thestring around a larger diameter pulley and dropped the mass holder,the acceleration of the system would be:

exactlyα? less than a? more than a?
Suppose that you measured themoment of inertia of the system as I = 5.05 kg-m2. Assume the diameterof the pulley around which you wrap the string is d =2.5 cm.

a) In trial 1:the system is initially spinning at angular speedωo = 9.27 rad/s. After you drop a disk of mass856 g on it, the system rotatesat ωf = 8.35 rad/s. Find:
- the moment of inertia of thedisk: kg-m2
- the radius of the disk: cm

b) In trial 3: suppose the system is initially spinning at angularspeed ωo = 2.05 rad/s. After you drop a cube of mass786 g on it, the system rotatesat ωf = 1.79 rad/s. Find:
- the moment of inertia of thecube: kg-m2
- the length of one side of the cube: cm

A playground ride consists of a disk of mass M = 56 kg and radius R= 2.0 m mounted on a low-friction axle. A child of mass m = 21 kgruns at speed v = 2.8 m/s on a line tangential to the disk andjumps onto the outer edge of the disk.



ANGULAR MOMENTUM
(a) Consider the system consisting of the child and the disk, butnot including the axle. Which of the following statements are true,from just before to just after the collision?
The momentum of the system doesn't change.
The momentum of the system changes.
The axle exerts a force on the system but nearly zero torque.
The torque exerted by the axle is zero because the force exerted bythe axle is very small.
The angular momentum of the system about the axle changes.
The angular momentum of the system about the axle hardlychanges.
The torque exerted by the axle is nearly zero even though the forceis large, because || is nearly zero.



(b) Relative to the axle, what was the magnitude of the angularmomentum of the child before the collision?
|C| = kg·m2/s
(c) Relative to the axle, what was the angular momentum of thesystem of child plus disk just after the collision?
|C| = kg·m2/s
(d) If the disk was initially at rest, now how fast is it rotating?That is, what is its angular speed? (The moment of inertia of auniform disk is ½MR2.)
= radians/s
(e) How long does it take for the disk to go around once?
Time to go around once = s
ENERGY
(f) If you were to do a lot of algebra to calculate the kineticenergies before and after the collision, you would find that thetotal kinetic energy just after the collision is less than thetotal kinetic energy just before the collision. Where has most ofthis energy gone?
Increased chemical energy in the child.
Increased thermal energy of the disk and child.
Increased translational kinetic energy of the disk.



MOMENTUM
(g) What was the speed of the child just after the collision?
v = m/s
(h) What was the speed of the center of mass of the disk just afterthe collision?
vcm = m/s
(i) What was the magnitude of the linear momentum of the disk justafter the collision?
|p| = kg·m/s
(j) Calculate the change in linear momentum of the systemconsisting of the child plus the disk (but not including the axle),from just before to just after impact, due to the impulse appliedby the axle. Take the x axis to be in the direction of the initialvelocity of the child.
px = px,f - px,i = kg·m/s

ANGULAR MOMENTUM
(k) The child on the disk walks inward on the disk and ends upstanding at a new location a distance R/2 = 1 m from the axle. Nowwhat is the angular speed? (It helps to do this analysisalgebraically and plug in numbers at the end.)
= radians/s
ENERGY
(l) If you were to do a lot of algebra to calculate the kineticenergies in part (k), you would find that the total kinetic energyafter the move is greater than the total kinetic energy before themove. Where has this energy come from?
Decreased chemical energy in the child.
Decreased momentum of the disk.
Decreased thermal energy of the disk and child.

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