Sample test 3, rotations and angular momentum
Nov, 2001
General Physics I

Table of moment of inertia (in case you need it)
	I disk = 1/2 M R^2
	I sphere = 2/5 M R^2
	I rod, center = 1/12 M L^2  , where L = total length of the rod
	I rod, end   = 1/3 M L^2

1)	Two balls, one massing 1 kg, and the other massing 2 kg, 
	are connected by a 2 meter string. 
	(The stringÕs mass is negligable.)  The objects are out in space, far
	from any gravitational forces or other influences.  
	The balls are turning about their center of mass at a rate of
	3 revolutions per second.

	(a) Where is the center of the mass of this system?

		Answer:
		1kg --------------- 2kg
		0m . . . . . . . . . . . . . . . . . . . 2m

		Given a coordinate system centered at the left, we have
		CM = ( 1kg * 0 m + 2k * 2m ) / (1 kg + 2 kg ) = 1.33 m

	(b) Find the  moment of inertia.  
		
		Answer:
		Around the center of mass, which is the axis of rotation, there 
		are two masses : 1 kg, with r=1.33m , and 2kg, with r=(2m-1.33m) = 0.66m

		So moment of inertia = I 
		 = sum( m r^2 ) = 1 kg (1.33m)^2 + 2kg(0.66m)^2 = 2.64 kg m^2

	(c) Find the kinetic energy of the system in two different ways
	     and show that they are equal:
	     A. Using the angular rotation speed and the total moment of 
	         inertia,
	
		Answer:
		The formula for rotational kinetic energy is 
		Kinetic Energy = 1/2 Mtotal Vcm^2 + 1/2 I omega^2
		Here Vcm = center of mass velocity = 0, so we ignore that part.
		omega = 2 Pi * 3 rev/sec = 6 pi radians / sec 
		so KE = 1/2 I omega^2 = 1/2 (2.64 kg m^2) (6 pi /sec)^2 = 470 Joules

	     B. Using the tangential speed of each ball, and adding their
                  tangential (linear) kinetic energies.

		Answer:
		The linear (tangential) velocity is V=(r omega).  We have to calculate this for
		each mass, using the same omega but different rÕs, 
		and then use KE = sum( 1/2 mass v^2).  This gives
		KE = (1/2) 1 kg ( 1.33 m 6 pi /sec )^2 + (1/2) 2 kg ( 0.66 m 6 pi/sec)^2
		= 470 Joules
		= same as part A.

2)	A flywheel (i.e. solid disk) masses 3 kg and has a radius of 0.25 m.
	Suppose it has a handle attached to the axle with a lever arm of
	1 meter, and you push on the end of the handle with a force of
	10 Newtons for 10 seconds.
	(a) How much torque are you applying?

		answer:
		Torque = radius * force (if perpendicular, which they are here)
		 = 10 Newtons *  1 meter = 10 newton meter
	(b) What is the angular acceleration of the disk?

		answer:
		Since torque = I * alpha 
		(i.e. torque = (moment of intertia) * (angular acceleration))
		then alpha = angular acceleration = torque / I .
		This is just like Ò a = F/mÓ for linear motion.
		We know the torque, and the moment of inertia for a solid disk is 
		I = 1/2 Mass Radius^2 = 0.5 3 kg (0.25 m)^2 = 0.094 kg m^2.
		So alpha = angular accel = ( 10 newton m ) / ( 0.094 kg m^2 ) 
		 = 106 radians/sec^2

	(c) How fast is the disk turning at the end of the 10 seconds?

		answer: 
		for linear motion,  Ò Vfinal - Vinitial = accel * timeÓ.
		for this rotational motion, Ò omegaFinal - omegaInitial = alpha*time Ò
		and omegaInitial=0.  So 
		omegaFinal = 106 rad/sec^2 * 10 sec = 10^4 rad/sec = 340 rev/sec
		

3)	A star spins around in about a week.  Suppose its hyrdrogen
	fuel runs out, and it shrinks into a white dwarf without losing any
	mass.  How long does it take to rotate after it shrinks?
	(R_star = 10^5 km , R_whitedwarf = 5 10^3 km).
	Explain what physics principles you are using to reach a solution.

		Answer:
		Since there arenÕt any outside forces, there are no outside
		torques and therefore angular momentum is conserved.
		But since the radius changes, the moment of inertia changes,
		and therefore the speed of rotation changes.  In particular,

		L  =  I omega =  (2/5 mass * R^2 )(2 pi / period) 
		 =  same before and after.
	
		And since the radius decreases by a factor of 20, 
		the moment of inertia I decreases by 20^2 = 400,
		and so the period also must decrease by a factor of 400.
		Thus the final period is (1 week / 400 ) 
		 = (24 * 60 min / 400) = 3.6 minutes.


4)	In class we set up the equations of motion for a rotating
	object rolling down an inclined ramp, solving the equations of linear
	and angular motion simultaneously.  Set up those equations, 
	including all the forces, and solve them to find the acceleration
	of a sphere down a ramp inclined at 45 degrees.

		Answer:

		The acceleration of gravity is g/sqrt(2) along the ramp,
		as well as g/sqrt(2) normal to the ramp
		( sin(45)=cos(45)=1/sqrt(2) )
		The friction force provides a torque to rotate the sphere,
		while the gravity force moves but does _not_ rotate it.
		(Gravity is effectively felt at the center of mass, and 
		 so there is no lever arm to give a torque.)

		So we have 
		linear equation:
		  mass g/sqrt - Friction = mass accel
		angular equation :
		  Friction * Radius =  I * alpha
		constraint equation needed for rolling without slipping:
		  radius * alpha = accel    
		sphere: 
		  I = 2/5 * mass * Radius^2

		and we solve all of these simultaneosly.

		The answer turns out to give an effective ÒinertiaÓ which
		is mass * (1 + 2/5), increased from just ÒmassÓ from
		the fact that the gravitational force has to rotate as well
		as move the sphere.  
		Thus the acceleration is  Ò g * 1/sqrt(2) * 1/(1 + 2/5) Ò
		  = 5 m/sec^2