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Unformatted text preview: Physics 16 – Spring 2010 – Problem Set 5 5.104. IDENTIFY: Apply SET UP: The block has acceleration to the block. The block moves in a horizontal circle of radius with the vertical. , so be to the left and let . . (b) gives . . The number of revolutions per second is (c) If , and . . . . Each . The freebody diagram for be upward. , directed to the left in the figure in the problem. string makes an angle the block is given in Figure 5.104. Let EXECUTE: (a) gives . The number of revolutions per minute is EVALUATE: The tension in the upper string must be greater than the tension in the lower string so that together they produce an upward component of force that balances the weight of the block. Figure 5.104 5.115. IDENTIFY: Apply to the circular motion of the bead. Also use Eq.(5.16) to relate to the period of rotation T. SET UP: The bead and hoop are sketched in Figure 5.115a. The bead moves in a circle of radius The normal force exerted on the bead by the hoop is radially inward. Figure 5.115a The freebody diagram for the bead is sketched in Figure 5.115b. EXECUTE: Figure 5.115b Combine these two equations to eliminate n: and so where T is the time for one revolution. so Use this in the above equation: This equation is satisfied by which gives (a) 4.00 rev/s implies Then and so this requires is not possible. So approaches so or by (b) This would mean But as the hoop rotates very fast, but (c) 1.00 rev/s implies The equation then says which is not This possible. The only way to have the equations satisfied is for means the bead sits at the bottom of the hoop. EVALUATE: so as (hoop moves faster). The largest value T can have is given by This corresponds to a rotation rate of For a rotation rate less than 1.58 rev/s, is the only solution and the bead sits at the bottom of the hoop. Part (c) is an example of this. 5.118. IDENTIFY: Apply to the car. It has acceleration , directed toward the center of the circular path. SET UP: The analysis is the same as in Example 5.24. EXECUTE: (b) (a) , where the minus sign indicates that the track pushes down on the car. The magnitude of this force is 30.4 N. EVALUATE: . . ...
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 Spring '10
 LorenzoSorbo
 mechanics, Acceleration

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