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**Hints and Examples in Chapter 10**

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Hints 10-10: t=r*F and t=Ia 10-32: Part a) See previous hint. b) w2=w20+2*a*q c) W=tq d) P=tw 10-40: I1=Ibody + Ithin rod at r=1.2 m and I2=Ibody + Ithin rod at r=25 cm and then use conservation of L 10-54: q= ½ at2 where s=rq

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10:15 A solid uniform cylinder with mass of 8.25 kg and diameter of 15 cm is spinning at 220 rpm on a thin frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of friction between the brake and rim is What must the normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

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**The total angular distance is 5.25 revolutions or 33 radians.**

w0=220 rpm=23 rad/s tq=W=-mk*n*R Iaq=W=-mk*n*R I= ½ *m*R2 = ½ (8.25)(0.075 m)2=0.023 kg*m2 w2=w20+2*a*q so –w02/2q=a=-8.04 rad/s2 n= .023*8.04*33/0.333/.075=7.47 N

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10:39 A small block on a frictionless horizontal surface has a mass kg. It is attached to a massless cord passing through a hole in the surface. The block is originally revolving at a distance of 0.3 m from the hole with an angular speed of 1.75 rad/s . The cord is then pulled from below shortening the distance the radius in which the block revolves to 0.15 m. Treat the block as a particle Is angular momentum conserved? What is the new angular speed? Find the change in kinetic energy of the block. How much work was done by pulling the cord?

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Part a) Since there is no torque acting on the particle (its angular velocity is not changing), the L is conserved. Part b) Let I=mr2 so I1w1=I2w2 so w2=w1*(r1/r2)2 w2=4*w1=7 rad/s Part c) DK= ½ I2w22-½ I1w21=1.03 x 10-2 J Part d) no other force does work so just change in KE

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10:43 A large wooden turntable in the shape of a flat disk has a radius of 2 m and total mass of 120 kg. The turntable is initially rotating at 3 rad/s about a vertical axis through its center. Suddenly a 70 kg parachutist makes a soft landing on the edge. Find the angular speed after the parachutist lands

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I1w1=I2w2 I1= ½ mr2=0.5 (120 kg)*(2)2= 240 kg*m2 w1=3 rad/s I2=I1+ m2r2= *22=520 kg*m2 3*240/520=w2=1.38 rad/s

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Chapter 10 Chapter 10 Rotational motion Rotational motion Part 2 Part 2.

Chapter 10 Chapter 10 Rotational motion Rotational motion Part 2 Part 2.

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