Question

A mass m = 0.6kg is attached to a string of length L = 1.5m. The string is pulled in such a way that it makes an angle of 25 degrees with the vertical direction as shown in the figure below. The mass is released from rest.

a. Find the speed of as the mass passes through the bottom of its trajectory.

Ubottom

b. What is the tension in the string when the mass passes through the bottom of its trajectory?

Answer 1

at the extreme position the body is at a height of h = L*(1-cos(25)) = 1.5*(1-cos(25)) = 0.14 m

apply law of conservation of energy

m*g*h = 0.5*m*v^2

v= sqrt(2*g*h) = sqrt(2*9.81*0.14) = 1.66 m/s

B) at the bottom

T + (m*v^2/L) = m*g

Tnesion T = m*(g- v^2/L )

T = 0.6*(9.81 - 1.66^2/1.5) = 4.78 N