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MasteringPhysics: Ch6 - Google Chrome @ https://session.masteringphysics.com/myct/itemView?assignmentProblemID 52314808 Fall2015 Ch6 Practice Problem 6.2 Signed in as Robert Jones Help Close Resources previous 5 of 50f 10 | next SOLUTION Practice Problem 6.2 SET UP (Figure 1) shows our diagrams. Two forces act on the ball: its weight w and the tension force FT exerted by the cord. In our free-body diagram, we point the r axis toward the center of the circle (in the direction of the centripetal acceleration ar d and we replace the tension force by its x and y components: Fz Fr sin and y Fr cos The y components of force sum to zero, so the net force on the ball is the x component of tension, which acts toward the center of the ball's circular path Consider a circular-motion problem in which tension not only must provide the proper radial acceleration but also must balance gravity. As we shall see, it can do both. A tetherball is attached to a swivel in the ceiling by a light cord of length L, as measured from the ceiling to the center of the tetherball. When the ball is hit by a paddle, it swings in a horizontal circle with constant speed u, and the cord makes a constant angle with the vertical direction. The ball goes through one revolution in time T. Assuming that T, mass m, and length L of the rope are known, derive algebraic expressions for the tension FT in the cord and the angle is known, we express ard in terms of T and the radius of the circle, using the following equation: ard 42R/T2 Note that the radius of the circular SOLVE Because the period path is L sin , not L.) To find Fr, we use Newton's second law for the r component of net force Next, we substitute the expression for the radius R of the circle (R-L sin ) into the preceding equation 4rVL sin ) Pr, sin = m The factor sin divides out, and we get We're halfway home; the foregoing equation gives the rope tension Fr in terms of known quantities. To obtain an expression for the angle B, we use Newton's second law for the y component of net force. (Remember that this component must be zero (Fy-0) because there is no y component of acceleration.) Newton's second law yields Figure of 1 We substitute our expression for Fr into this equation to get 4T2 L So we've succeeded in expressing the tension Fr and the cosine of the angle in terms of known quantities REFLECT For a given length L. cos decreases and increases as T becomes smaller and the ball makes more revolutions per second. The angle can never be 900, however; that would require that T 0 Fr oo and u oo. The relationship between and doesn't involve the mass of the ball, but the force magnitude is directly proportional to m. When is very small. Fr is approximately equal to the ball's weight, cos is approximately unity, and the expressions we found above give the approximate expression T-2 V L/g) Part A - Practice Problem aad What is the period T of the ball's motion if 4.0 m and the cord makes an angle of -40° with the vertical? Express your answer in seconds to two significant figures (a) The situation (b) The forces on the ball (c) Free-body diagram of the ball Ask me anything 8:58 PM 10/13/2015

Explanation / Answer

Here ,

length of cord , L = 4 m

angle , B = 40 degree

Now , as cos B = g * T^2/( 4 *pi^2 * L)

cos(40) = 9.8 * T^2 /( 4 * 3.141^2 * 4)

solving for T

T = 3.51 s

the time period of motion is 3.51 s

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