Rotational Part IV – Angular

Pre AP and Its Conservation In an analogy with linear momentum, we can define angular momentum L: L = I    Since  = I and since  fi , We can then t write the total as being the of change of angular momentum.    fi I fi    L I     t tt

If the net torque on an object is zero, the total angular momentum is constant. Angular Momentum and Its Conservation Another way of expressing the angular momentum L of a relative to a point O is the of the particle's r relative to O with the linear momentum p of the particle.      L  r  p  r mv

Angular momentum of a particle

The value of the angular momentum depends on the choice of the origin O, since it involves the position vector relative to the origin The units of angular momentum: kgm2/s NOTE: If  = v/r and I = mr2, then L = mr2 OR L = mvr Change in angular momentum

τ  (t) = Iωf – Iωi τ = torque t = I = of

ωf = final angular

ωi = initial

The angular momentum of a system remains unchanged unless an external torque acts on it Spinning ice skater Arms extended, inertia is larger, velocity is smaller Pull arms in tight, inertia decreases, velocity increases Therefore, systems that can change their rotational inertia through internal will also change their rate of : Example #1 If the ice cap at the South Pole melted and the water were uniformly distributed over the 's oceans, what would happen to the earth’s angular velocity? Would it increase, decrease or stay the same? REASONING AND SOLUTION Consider the earth to be an isolated system. Note that the earth rotates about an axis that passes through the North and South poles and is to the of the equator. If the ice cap at the South Pole melted and the water were uniformly distributed over the earth's oceans, the at the South Pole would be uniformly distributed and, on average, be farther from the earth's rotational axis. The of the earth would, therefore, increase. Example #1 con’t Since the earth is an isolated system, any involved in the redistribution of the water would be internal torques; therefore, the angular momentum of the earth must remain the same. If the moment of inertia increases, and the angular momentum is to remain constant, the angular velocity of the earth must decrease.

Rotational Inertia, I, INCREASES BUT angular momentum, L, stays the same  Since L = I Angular velocity, , DECREASES Example #2 – Pulling Through a Hole

A particle of mass m moves with vi in a of ri on a frictionless tabletop. The particle is attached to a massless string that passes through a hole in the table as shown. The string is pulled slowly

downward until the particle is a rf from the hole and continues to rotate in a circle of that radius. (A) Find

the final velocity vf. (B) Find the T in the string when the particle moves in a circle of radius r in terms of the angular momentum L.

vrii v f  (A) Lf = Li  mv r = mv r f f i i rf

T = centripetal

2 mv2 m L L2 (B) T  r r mr L = mvr  v = L/mr mr3 Example #3 A skater has a moment of inertia of 3.0 kg m2 when her arms are outstretched and 1.0 kg m2 when her arms are brought in close to her sides. She starts to at the rate if 1 rev/s when her arms are outstretched, and then pulls her arms to her sides. (A) What is her final angular speed? (B) How much did she do?  Ii (A) fi   Lf = Li  Iff = Iii  I f 2 3 kg m 3 rev/sec f 2 1rev / sec 1 kg m (B) Work (rotational) = KE (rotational)  Work- Theorem

1122 Work I     2 2 22f f i i W = (.5)(1.0)(3 x 2π) – (.5)(3)(1 x 2π) 118 J Angular Momentum Angular momentum depends on linear momentum and the distance from a particular point. It is a vector quantity with symbol L. If r and v are  then the of angular momentum with respect to point Q is given by L = p∙r = m v r. In the case shown below L points out of the page. If the mass were moving in the opposite direction, L would point into the page. The SI unit for angular momentum is the kg  m2 / s. (It has v no special name.) Angular momentum is a . A torque is r m Q needed to change L, just a force is needed to change p. Anything spinning has angular has angular momentum. The more it has, the harder it is to stop it from spinning. If r and v are not  then the between these two vectors must be taken into account. The general definition of angular momentum is given by a vector cross product:      L  r  p  r mv

This formula works regardless of the angle. The magnitude of the angular momentum of m relative to point Q is: L = r p sin = m v r. v

 r m

Q Now if a acts on an object, it must accelerate, which means its momentum must change. Similarly, if a net torque acts on a body, it undergoes angular , which means its angular momentum changes. Recall, angular momentum’s magnitude is given by if v and r are L = mvr mutually perpendicular So, if a net torque is applied, angular velocity must change, which changes angular momentum. So net torque is the rate of change of angular momentum, just as net force is the rate of change of linear momentum. Here is yet another pair of similar equations, one linear, one rotational. From the formula v = r , we get L = mvr = m r (r ) = m r2  = I  This is very much like p = mv, and this is one reason I is defined the way it is. In terms of magnitudes, linear momentum is inertia speed, and angular momentum is rotational inertia times angular speed.

NOTE: If  = v/r and I = mr2, then L = mr2 OR L = mvr Suppose Mr. Stickman is sitting on a stool that swivels holding a pair of dumbbells. His axis of rotation is vertical. With the far from that axis, his moment of inertia is large. When he pulls his arms in as he’s spinning, the weights are closer to the axis, so his moment of inertia gets much smaller. Since L = I  and L is conserved, the product of I and  is a constant. So, when he pulls his arms in, I goes down,  goes up, and he starts spinning much faster.

I  = L = I Since an external net force is required to change the linear momentum of an object. An external net torque is required to change the angular momentum of an object.

It is easier to balance on a moving bicycle than on one at rest. • The spinning wheels have angular momentum. • When our center of is not above a point of support, a slight torque is produced. • When the wheels are at rest, we fall over. • When the bicycle is moving, the wheels have angular momentum, and a greater torque is required to change the direction of the angular momentum. Angular momentum is conserved when no external torque acts on an object.

Angular momentum is conserved for systems in rotation. The law of conservation of angular momentum states that if no unbalanced external torque acts on a rotating system, the angular momentum of that system is constant. With no external torque, the product of rotational inertia and rotational velocity at one time will be the same as at any other time. is controlled by variations in the body’s rotational inertia as angular momentum is conserved during a forward somersault. This is done by moving some part of the body toward or away from the axis of rotation. Example #4 What is the angular momentum of a 0.210-kg ball rotating on the end of a thin string in a circle of radius 1.10 m at an tangential speed of 5.25 m/s? m = 0.210 kg; r = 1.10 m; v = 5.25 m/s

L = mvr L = (0.210 kg)(5.25 m/s)(1.10 m) L = 1.21 kg∙m2/s Example #5 What is the angular momentum of a 0.210-kg ball rotating on the end of a thin string in a circle of radius 1.10 m at an angular speed of 10.4 rads/sec? m = 0.210 kg; r = 1.10 m;  = 10.4 rads/sec

L = I = mr2

L = (0.210 kg)(1.10 m)2(10.4 rads/s)

L = 2.64 kg-m2/s Example #6 What is the angular momentum of a 2.8-kg uniform cylindrical grinding wheel of radius 18 cm when rotating at 1500 rpm? m = 2.8 kg; r = 0.18 m;  = 1500 rpm L = I 1 2 Icylinder  2 mr 2 1500rpm= 50 rads/sec 60

1 2 L  2 2.8 kg 0.18 m  50 rads / sec

L = 7.1 kg-m2/s Example #7 An artificial is placed into an elliptical about the earth. Telemetry data indicate that its point of closest 6 approach (called the perigee) is rP = 8.37 × 10 m from the center of the earth, and its point of greatest distance (called 6 the apogee) is rA = 25.1 × 10 m from the center of the earth. The speed of the satellite at the perigee is vP = 8450 m/s. Find its speed vA at the apogee.

LApogee = LPerigee  IAA = IPP

22vvAP    mrAP    mr    rrAP   

8.37 106 m 8450 m / s rvPP    v   vA  A 25.1 106 m rA 2820 m/s Example #8 A horizontal disk of rotational inertia 4.25kg ⋅ m2 with respect to its axis of is spinning counterclockwise about its axis of symmetry, as viewed from above, at 15.5 revolutions per on a frictionless massless bearing. A second disk, of rotational inertia 1.80 kg ⋅ m2 with respect to its axis of symmetry, spinning clockwise as viewed from above about the same axis (which is also its axis of symmetry) at 14.2 revolutions per second, is dropped on of the first disk. If the two disks stick together and rotate as one about their common axis of symmetry, at what new angular velocity would the combined disks move in rads/sec? Example #8 Continued

15.5revs 2 rads i  = +31 rads/s (counterclockwise) sec rev 14.2revs 2 rads = −28.4 rads/s (clockwise) i   sec rev

L Before = L After

I11 + (−I22) = I1Final + I2Final Example #8 Continued

I11 + (−I22) = I1Final + I2Final

II1 1 2 Final  II12

4.25kg m22 31 rads / s  1.80 kg  m 28.2 rads / s   Final 4.25kg m22  1.80 kg  m

Final = 41.9 rads/s The Vector Nature of Rotational

The vector directions of the angular velocity vector  and the angular momentum vector L  are along the axis of rotation. Applying the right-hand rule gives the direction. Right-Hand Rule Grasp the axis of rotation with your right hand, so that your fingers circle the axis in the same sense as the rotation. Your extended thumb points along the axis in the direction of the angular velocity vector.

Angular acceleration arises when the angular velocity changes, and the acceleration vector also points along the axis of rotation. The acceleration vector has the same direction as the change in the angular velocity. and angular momentum vectors also point along the axis of rotation. Example #9 Tipping the Wheel A student sitting on a stool on a frictionless turntable is holding a rapidly spinning bicycle wheel. Initially, the axis of the wheel is horizontal, with the angular momentum vector L pointing to the right. What happens when the student tips the wheel so that the spin axis is vertical, with the wheel spinning counterclockwise? Answer The system is free to rotate about the vertical axis (no vertical torques) and initially the angular momentum is zero along that axis. Therefore, vertical angular momentum is conserved and the final angular momentum must also be zero. In the final state, the wheel has a large angular momentum pointing vertically upward, so the stool and student must rotate clockwise to have an equal and opposite downward angular momentum.