Reality Shows Essay

ROTATIONAL MOTION PROBLEMS: 09-1 1) A grinding wheel starts from rest and has a constant angular acceleration of 5 rad/sec2. At t = 6 seconds find the centripetal and tangential accelerations of a point 75 mm from the axis. Determine the angular speed at 6 seconds, and the angle the wheel has turned through. |We have a problem of constant angular acceleration. The figure & coordinate system are |[pic] | |shown. Since a time is given in the problem we must use the equations of motion. | | | | |((t) = (1/2) ( t2 + ( 0 t ; ((t) = ( t + ( 0 | | | | | |The initial state of motion is: ( 0 = 0; ( 0 = 0 . We are given ( in the problem. | |Hence we have the specific equations of motion: | | ((t) = (1/2)(5) t2 ; ((t) = 5 t . Thus at time t = 6 sec, ((6 sec) = (5)(6) = 30 rad/sec . And: ((6 sec) = (1/2)(5)(6)2 = 90 rad . The centripetal & tangential accelerations are linear quantities. Hence, to calculate we need the interconnecting equations. ( s = r (( ; v = r ( ; a t = r ( Where all angular quantities are expressed in terms of angular units of radians. ) Thus, at 6 seconds we have: for velocities for acceleration v(6 sec) = r ( = (. 075)(30) = 2. 25 m/sec a t = r ( = (. 075)(5) = . 375 m/sec2 a c = v2/r = r ( 2 = (. 075)(30)2 = 67. 5 m/sec2 . The magnitude of the total linear acceleration of this point would be given by: [pic] 2) An electric motor operates at 1800 rpm. Find its angular speed in radians/second.

The torque |[pic] | |which produces this rotation must therefore be a clockwise (‘in’) torque. Thus, ( and (( | | |are in the same direction (they both point into the paper). Thus we will have: | | | | | |W(by net ( ) = ( net (( = (150)(1875) = 2. 1 x 105 J. | | | | | |The final kinetic energy (in rotational form) can be calculated thusly: | | KEfinal = (1/2) I ( 2 = (. 5)(25)(150) = 2. 81 x 102 J. We note that this is the same as the work done by the ‘net torque’. This is as expected, since we have: translational: W(by net force) = ( KE(translational) rotational: W(by net torque) = ( KE(rotational) . ) A solid sphere with a radius of 75 mm rolls (without slipping) down an inclined plane that is 5 m long. What is the angular velocity of the sphere at the bottom of the plane if it requires 10 seconds for it to reach the bottom? What is the angle of the plane? |Solution: We note that a time is given in the problem. Hence we must use the |[pic] | |equations of motion. To see where they will enter, we will take a W-E | | |approach to the problem.

The figure is drawn, and we select Ug = 0 at the | | |base. The work energy theorem is: | | | | | |Wby me = ( KE + ( U + Wancf . | |

Since the ball rolls without slipping then no work is done by, or against friction & Wancf ( 0. I clearly do no work, so we have a problem of Conservation of total Mechanical Energy. Thus: KEtop + Utop = M g h = KEbottom + Ubottom = (1/2)M v2 + (1/2) I ( 2 . Note that the ball possesses both types of kinetic energy at the bottom. Thus: M g h = (1/2)M v2 + (1/2)(2/5)M R2(v/R)2 ( g h1 = {(1/2) + (1/5)} v2 = (7/10)v2 . Thus: [pic] . (1) If the angle of the plane were known, then we could calculate ‘h’ and determine the speed at the bottom. 09-4 Associated with the ‘bottom’ we know the given quantity t = 10 seconds. This is where the |[pic] | |equations of motion come into play. Let us construct the equations of motion (translational) | | |for the CM of the ball. Choosing a CS as shown we have: | | | | | |x(t) = (1/2) a t2 ; v(t) = a t . | At t= 10 sec, x (10sec) = 5 m = (1/2) a (10)2, or a = 0. 1 m/sec2 . Thus the linear speed at the bottom is: v = (. 1)(10) = 1 m/sec. Since this is the same as the linear speed of a point on the surface of the sphere (it rolls without slipping), then the angular speed at the bottom is ( = v/r = (1 m/s)/(. 075 m) = 13. 3 rad/sec. Using the result for v in equation (1) above, we calculate ‘h’. h = (7/10) v2/g = (. 7)(1)2/(9. 8) = . 0714 m .

The angle of inclination of the plane is then determined as follows: sin ( = h/(5m) = (. 0714)/(5) = . 01428 From a calculator this gives: ( = 0. 818 degrees. However, we note that since the angle is small, then we can use the approximation: (For small ( – in radians): sin ( ( ( . Thus we could state the answer as: ( ( . 01428 radians. |5) Assume that you are given the following information about the system shown: m1, m2, |[pic] | |M, R, I, ( , ( k . | | | | |When the system is released, block #1 accelerates downward. Apply Newton’s 2nd law to each of| | |the objects in the system in order to develop a set of equations that could be solved | | |yielding the acceleration of the system.

Be certain to identify your ‘known’ and ‘unknown’ | | |quantities. | | 09-5 |Solution: We have a mixed translational and rotational problem. Blocks ‘A’ & ‘B’ |[pic] | |undergo translational motion (1-dim. ), while the pulley (our 3rd real object) | | |undergoes rotational motion.

Let’s start with block ‘A’: | | | | | |For ‘B’: ( Fx: m B g – T2 = m B a (1) | | | | | |For ‘A’: ( Fx: T1 – m A g sin ( – f k = m A a (2) | | | | | |( Fy: N A – m A g cos ( = 0 (3) ; f k = ( k N A (4) | | At this point our list of ‘unknowns’ includes: a, T1, T2, N, f k . We need another equation. |For pulley: |[pic] | | | | |( ( cw: T2 R – T1 R = I ( (5) | | | | | |This gives us a 5th equation, but unfortunately also introduces a ‘new unknown’, ( . | The solution of our difficulty rests in recognizing that the linear motions of the blocks and the rotational motion of the pulley are inter-connected. Hence, we have, for any point on the outside edge of the pulley, ( s = r (( ; v = r ( ; a t = r ( If the rope does not slip on the pulley, then the linear acceleration of any point on the rope is the same as at for the outside edge of the pulley. But the linear acceleration of the rope is the same as that of the two blocks. Thus: a t = a = R ( (6) Unknowns are: a, T1, T2, N, f k, ( .

Thus six equations in six unknowns guarantees a solution. 6) A pulley has mass: m = 1 kg, radius: R = 10 cm, and a moment of inertia about its center of 0. 8 m R2. It has a cord wrapped around it with one end fastened to the ceiling. The pulley wheel is released from rest and falls a distance ‘h’. (a) Consider the motion of the center of mass. Apply Newton’s 2nd law to the translational & rotational motions, and calculate the angular acceleration of the wheel, and the tension in the cord. (b) Find the speed ‘v’ of the CM of the wheel after it has fallen a distance ‘h’ (let ‘h’ = 1 meter). 09-6 |Solution: The figure is as shown, and we have all forces acting on the wheel.

The |[pic] | |motion of the CM is governed by Newton’s 2nd law: | | |( Fy: m g – T = m a (1) | | | | | |The rotation of the object about its CM is governed by the rotational 2nd law: | | | | | |( ( ccw: T R = I ( (2) | |

The two motions are not independent. They are ‘inter-connected’ through the relations: ( s = r (( ; v = r ( ; a t = r ( Here ‘v’ represents the linear speed of a point on the outside of the wheel. This is the same as the linear speed vcm of the CM. The acceleration of the CM, a cm , is the same as a t , the tangential acceleration of a point on the outside rim of the wheel. Substituting for ( in equation (2), then gives: m g – T = m a (a) T R = (. 8) m R2 (a / R) ( T = (. 8) m a (b) Eliminating T gives: m g = (1. 8) m a ( a = g/(1. 8) = 5. 44 m/sec2.

Then ( = a /R = (5. 44)/(. 1) = 54. 4 rad/sec2 . & T = . 444 m g = 4. 35 N. If we construct equations of motion for the CM ( yo = 0; vo = 0), we have: y(t) = (1/2) a t2 = (1/2)(5. 44) t2 ; v(t) = a t = 5. 44 t . When the wheel has ‘fallen’ a distance ‘h’, h = (1/2)(g/1. 8) t’2 . This yields: [pic] Now we note that 1. 8 = 18/10 = 9/5 so that we can write: [pic] . Numerically this gives: 3. 3 m/sec. Consider now this last equation for v(t’). It is somewhat familiar. We recall that if an object ‘falls’ (accel. = g) through a height ‘h’, then its speed is given by [pic] . Where does the factor: (5/9) come from?

Let us calculate the speed via work energy considerations. We have only two forces present: gravity ( Ug; and 09-7 the tension ‘T’. What becomes of the work done by the torque produced by T? This is converted into rotational kinetic energy. Hence the W-E theorem becomes: W by me = ( KE trans + ( KE rot + ( U + W ancf . Thus, since the initial KE = 0, and U = – m g h, then we have: KEfinal = (1/2) m v2 + (1/2) I ( 2 = m g h . Substituting: ( = v/R and I = (. 8) m R2 we have: m g h = (1/2) m v2 + (1/2)(. 8) m R2 (v/R)2 . The M cancels, and we have: g h = (1/2) v2 + (1/2)(8/10) v2 = (1/2){ 1 + 4/5 } v2 = (1/2)(9/5) v2 .

Thus we obtain the factor of 5/9 (rather than 1) since the potential energy in this problem is converted into kinetic energy of both a translational, and a rotational nature. 7) A 25 kg boy stands 2 m from the center of a frictionless playground merry-go-round which has a moment of inertia of 200 kg-m2. If the boy begins to run in a circular path with a speed of 0. 6 m/sec relative to the ground, calculate: (a) the angular velocity of the MGR and (b) the speed of the boy relative to the surface of the MGR. Solution: In order for the boy to acquire a velocity he must be accelerated. The force that accomplishes this is friction. We have a problem in which forces of an unknown magnitude act between the boy & the MGR. To eliminate these ‘unknown’ forces (& resulting unknown torques) we take a systems approach. Initially the system has zero angular momentum. Suppose there is a radial line painted on the |[pic] | |MGR with an ‘X’ where the boy initially is standing. As the boy runs ccw through an angle (, | | |the MGR rotates cw by an angle (. | | | | | |Since there are no other forces but friction which can produce the torque which results in the | | |angular acceleration of the MGR (why can’t boy’s weight contribute? , then by treating the MGR | | |& boy together as one system we eliminate these ‘internal’ torques, and we have conservation of | | |angular momentum. | | L I = L f ( I b ( bi ( I M ( Mi = I b ( bf ( IM ( Mf Now: ( bi = ( Mi = 0 or L bf = – L Mf . Hence: m r2 ( bf = m r2 (v/r) = (25)(2)(. 6) = 30 kg-m2/s . 09-8 Then L Mf = IM ( Mf = 200 ( Mf = 30 . Thus ( Mf = 200/30 = . 15 rad/sec. Consider now the linear speed of a point on the MGR under the boy (that is, at a distance of 2 m from the center). This is: vM = r ( M = (2)(. 15) = 0. 3 m/sec. Since at any instant of time the point directly under the boy has a linear velocity of 0. m/sec in a direction opposite to the direction the boy is moving, then the speed of the boy relative to the MGR is: 0. 6 + 0. 3 = 0. 9 m/sec. We might also ask the question: “When will the boy return to his starting position (the ‘X’)? ” We can solve this using either angular variables or linear variables. Consider the latter. In a time (t the boy will move a distance ( s b = (. 6) (t relative to the ground. In the same time the ‘X’ will move a distance (s X = (. 3)t . The boy will return to the ‘X’ if the sum of these two equals the circular distance around at a radius of 2m. That is, (. 6) (t + (. 3) (t = 2 ( r = (2)(3. 14)(2) ( (t = 13. 96 sec. You will obtain the same result if you used the boy’s velocity relative to the MGR.