A simple pendulum consists of an object suspended by a string. Theobject is assu
ID: 1725151 • Letter: A
Question
A simple pendulum consists of an object suspended by a string. Theobject is assumed to be a particle. The string, with its top endfixed, has negligible mass and does not stretch. In the absence ofair friction, the system oscillates by swinging back and forth in avertical plane. If the string is 2.05 m long and makes an initial angleof 25.5° with the vertical,calculate the speed of the particle at the following positions. (a) at the lowest point in its trajectory(b) when the angle is 15.0°
(a) at the lowest point in its trajectory
(b) when the angle is 15.0°
Explanation / Answer
When the pendulum swings, the particle isdisplaced vertically. The vertical displacement is given by h = L* ( 1 - cos ) L = Lengthofstring = 2.05 m, = Anglebetween string and vertical Atextreme = 25.50 h = 2.05* ( 1 - cos25.50) = 0.20 m P.E. at extremeposition U = m* g * h = m * 9.8 *0.20 = 1.96 *m J a. When the pendulum is atloweastposition, h = 0 P.E. = 0 i.e. all the p.e. isconverted into k.e., hence (1/2) * m *v12 = U 0.5 * m *v12 = 1.96*m => speed v1 = (2 * 1.96) = 1.98 m/s b. at = 15.00 P.E. U1 = m* g * h1 = m * 9.8 *2.05 * (1 - cos15.00) = 0.68 *m Changeinp.e. = U - U1 = 1.96* m - 0.68 *m = 1.28 * m (1/2) *m *v22 = 1.28* m speed v2 = (2 * 1.28) = 1.60 m/s P.E. U1 = m* g * h1 = m * 9.8 *2.05 * (1 - cos15.00) = 0.68 *m Changeinp.e. = U - U1 = 1.96* m - 0.68 *m = 1.28 * m (1/2) *m *v22 = 1.28* m speed v2 = (2 * 1.28) = 1.60 m/s
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