Question

A pendulum consists of a 1.9-kg bob attached to a light 2.3-m-long string. While hanging at rest with the string vertical, the bob is struck a sharp horizontal blow, giving it a horizontal velocity of 4.1 m/s. At the instant the string makes an angle of 30° with the vertical, calculate the following:

speed= 3.28 m/s

gravitational potential energy is 5.739 J

what is the tension in the string in Newtons?

what is the angle of the string with the vertical when the bob reaches its greatest height?

Answer 1

(a)

The vertical height (h) at 30*
h = L - Lcosθ
h = 2.3[1-cos30*]
h = 0.308 m
Let the velocity is (v) of the bob at 30*
By the law of energy conservation:-
KE(initial) = KE(final) + PE(final)
1/2mu^2 = 1/2mv^2 + mgh
v^2 = u^2 - 2gh
v^2 = (4.1)^2 - 2 x 9.81x 0.308
v = √10.7670
v = 3.28 m/s
(b)

By PE = mgh
PE = 1.9 x 9.81 x 0.308
PE = 5.74 J
(c)

By T = mgcosθ*

T = 1.9 x 9.81 x cos30
T = 16.14 N

(d)

Let the maximum height attained is H
=>By the law of energy conservation:-
=>KE(initial) = PE(final)
=>1/2mu^2 = mgh
=>H = u^2/2g
=>H = (4.1)^2/(2 x 9.8)
=>H = 0.8576
=>By h = L - Lcosθ
=>H = 2.3[1-cosθ]
=>0.8576/2.3 = [1-cosθ]
=>cosθ = 0.6271
=>θ ≈ 51*