Physics 2001 Problem Set 5 Solutions

Size: px

Start display at page:

Download "Physics 2001 Problem Set 5 Solutions"

Andra Smith
6 years ago
Views:

1 Physics 2001 Poblem Set 5 Solutions Jeff Kissel Octobe 16, A puck attached to a sting undegoes cicula motion on an ai table. If the sting beaks at the point indicated in the figue, which path (A, B, C, o D) is the puck going to follow? You ae looking down at the table fom above. Figue 1: A ovehead view of a puck in unifom cicula motion. The velocity vecto in a cicula path always points tangential to the cicle. Thus, if instantaneously eleased (i.e. the sting beaks) the puck will have a velocity tangential to the cicle; the puck will tavel along path C. If the speed of the puck is 1.2 m/s, and the tension in the 1.0 m length sting is 35N befoe it beaks, what is the mass of the puck? Fo unifom cicula motion as we have hee, the centipetal acceleation is a C = v2 (1) which means the centipetal foce is F C = ma C = m v2 (2) 1

2 Jeff Kissel, Octobe 16, 2006 Physics 2001, Poblem Set 5 Solutions While the puck is still attached, in ode fo the puck to continue along the cicula path, the tension of the sting must balance the centipetal foce as seen in Fig. 1. We e given the tension T, the speed of the puck v, and the length of the sting (i.e. the adius of the cicle), so we can solve Eq. 2 fo the mass of the puck m. T = F C T = m v2 m = T v 2 (3) (35 N)(1 m) = (1.2 m s 1 ) 2 = kg (4) 2. A 0.2 kg flying pig toy is attached to the ceiling with a sting. When the pig s wings flap, it moves at a constant speed of 1.21 m/s in a hoizontal cicle of adius 0.44 m. Find (a) the angle the sting makes with the vetical and (b) the tension in the sting. Figue 2: A flying pig attached to the ceiling by a sting following a constant cicula path. Because the pig is following this cicula path at a constant speed, the sum of foces in any given instant in time on the path is zeo. As usual, thee ae two unknowns in the poblem, which we can find by summing the net foces in the ŷ and ˆx diections. 2

3 Jeff Kissel, Octobe 16, 2006 Physics 2001, Poblem Set 5 Solutions Fy 0 = T y mg T cosθ = mg T = mg cosθ Fy (5) 0 = T x F C F C = T sin θ ( ) m v2 mg = sin θ cosθ ( ) v 2 = sin θ g cosθ = tanθ ( ) v θ = tan 1 2 and plugging this back in Eq. 5, g (6) ( (1.21 m s = tan 1 1 ) 2 ) (0.44 m)(9.8 ms 1 ) = (7) T = (0.2 kg)(9.8 m s 1 ) cos(18.75 ) T = N (8) 3. Calculate you appaent weight at the top and bottom of a Feis wheel, given that the adius of the wheel is 7.2 m, it completes one evolution evey 12 s, and you mass is 55 kg. This poblem asks fo one s appaent weight, but it s bette think of it as you net acceleation. At both the top and bottom of the wheel one will feel the weight of gavity, mg. Howeve, the centipetal acceleation fom the tuning Feis wheel will change the net acceleation. The top and bottom will be diffeent because of the diection in which the centipetal acceleation is affecting the net acceleation. 3

4 Jeff Kissel, Octobe 16, 2006 Physics 2001, Poblem Set 5 Solutions We know the fomula fo centipetal acceleation involves the velocity of the cat v, but it s not explicitly given. We can figue it out fom the fequency. The distance taveled by the cat in one evolution, d is the cicumfeence of the Feis wheel, d = 2π. The poblem says that it completes this distance once evey 12 seconds, so the velocity of the cat is v = d t = 2π f (9) whee I ll plug in f as 1/12 s 1, o 1/12 Hz. So, in the instant we measue ou weight at the top of the Feis wheel, the only foces on use ae centipetal acceleation, a C in the positive ŷ diection, and ou weight, mg in the negative ŷ diection. So, F (T) net = F C W m net = ma C mg net = v2 g = (2πf)2 g net = 4π 2 f 2 g (10) = 4π 2 (7.2 m)( 1 12 Hz)2 (9.8 m s 2 ) net = m s 2 (11) Appaently, one feels lighte at the top of the Feis wheel! At the bottom, the centipetal acceleation is now in the same diection as gavity, so F (B) net = F C W 4

Jeff Kissel, Octobe 16, 2006 Physics 2001, Poblem Set 5 Solutions m net = ma C mg net = v2 g = (2πf)2 g net = 4π 2 f 2 g (12) = 4π 2 (7.2 m)( 1 12 Hz)2 (9.8 m s 2 ) net = 11.

5 Jeff Kissel, Octobe 16, 2006 Physics 2001, Poblem Set 5 Solutions m net = ma C mg net = v2 g = (2πf)2 g net = 4π 2 f 2 g (12) = 4π 2 (7.2 m)( 1 12 Hz)2 (9.8 m s 2 ) net = m s 2 (13) One feels heavie than nomal at the bottom! 4. Quiz: What is the obital speed of the Intenational Space Station? It is m fom the cente of the eath. In fact, the definition of space obit is when a satellite s centipetal acceleation matches it s acceleation due to gavity. So, we can as usual instantaneously sum the foces in the ŷ diection at any point in the obit. It will be in equilibium in this diection, because we don t want the ISS to float off into space, o cash back down to Eath (whose symbol is ). m v2 ISS 0 = F C F g = m GM 2 viss 2 = GM GM v ISS = ( = 11 N m 2 kg 2 )( kg ) ( m) v ISS = m s 1 (14) v ISS = km s 1 (15) 5

PS113 Chapter 5 Dynamics of Uniform Circular Motion

PS113 Chapter 5 Dynamics of Uniform Circular Motion PS113 Chapte 5 Dynamics of Unifom Cicula Motion 1 Unifom cicula motion Unifom cicula motion is the motion of an object taveling at a constant (unifom) speed on a cicula path. The peiod T is the time equied