Question

A 440 g mass is whirled in a circle at the end of a string of length L = 82 cm in a horizontal circle. The string makes an angle of 65° with the vertical. What is the speed of the mass?

Answer 1

given
m = 440 g = 0.44 kg
L = 82 cm = 0.82 m
theta = 65 degrees

radius of the circular path, r = L*sin(theta)

= 0.82*sin(65)

= 0.743 m

let T is the tension in the string.

here the mass is in equilibrium in y-direction.

so, Fnety = 0

T*cos(theta) - m*g = 0

T = m*g/cos(theta)

= 0.44*9.8/cos(65)

= 10.2 N

net force acting towards the center, Fnetx = T*sin(theta)

m*a = T*sin(theta)

m*v^2/r = T*sin(theta)

v^2 = r*T*sin(theta)/m

v = sqrt(r*T*sin(theta)/m)

= sqrt( 0.743*10.2*sin(65)/0.44)

= 3.95 m/s <<<<<<<<----------------Answer