A pendulum consists of a 1.8-kg bobattached to a light 2.5-m-long string.While h
ID: 1726067 • Letter: A
Question
A pendulum consists of a 1.8-kg bobattached to a light 2.5-m-long string.While hanging at rest with the string vertical, the bob is struck asharp horizontal blow, giving it a horizontal velocity of4.1 m/s. At the instant the stringmakes an angle of 30° with thevertical, calculate the following: (a) the speed1 m/s
(b) the gravitational potential energy (relative to its value is atthe lowest point)
2 J
(c) the tension in the string?
3 N
(d) What is the angle of the string with the vertical when the bobreaches its greatest height?
4° (a) the speed
1 m/s
(b) the gravitational potential energy (relative to its value is atthe lowest point)
2 J
(c) the tension in the string?
3 N
(d) What is the angle of the string with the vertical when the bobreaches its greatest height?
4°
Explanation / Answer
Mass of the bob, m = 1.8 kg Length of the string, L = 2.5 m Initial velocity at the lowest point, V = 4.1m/s Angle made with vertical in new position, = 30 degrees Height raised, h = L ( 1 - Cos ) = 2.5 ( 1 - 0.866 ) = 2.5 * 0.134 = 0.335 m (a) Kinetic energy at the lowest point = ( K.E + P.E) at the new position ( 1 / 2 ) m V 2 = ( 1 / 2 ) m V'2 + m g h V 2= V' 2 + 2 g h Velocity at the new position, V' = ( V2 - 2 g h ) = [ ( 4.1 )2- ( 2 * 9.8 * 0.335 ) ] = 3.2 m/s (b) Gravitational potential energy, PE = m g h = 1.8 * 9.8 * 0.335 = 5.9 J (c) Tension , T = ( m V 2 / L ) + m g Cos = ( 1.8 * 3.2 2 / 2.5 ) + 1.8 * 9.8 * 0.866 = 7.37 + 15.28 = 22.65 J (d) P.E at the highest point = K.E at the lowestpoint m g L ( 1 - Cos ) = ( 1/2 ) mV 2 g L ( 1 - Cos ) = ( 1/2 ) V2 9.8 * 2.5 * ( 1 - cos )= ( 1/2 ) * 4.1 2 24.5 ( 1 - cos ) = 8.41 1 - cos = 0.3431 cos = 0.6569 Angle made at the highest point, =48.9o
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.