Question

A cylindrical drum of mass M, radius R and moment of inertia Icm can rotate about its central axis. A Mass m is suspended from a light rope that has previously been wound around the drum. The mass is frees from the rest at a height h on the floor, unwinding the rope when falling. a) Ignoring the friction, use the Newton's second law to calculate the tension in the string while m is falling. b) Ignoring the friction, uses energy conservation to determine the speed of m when it reaches the floor. c) If M = 0.440 kg, R = 4.80 cm, m = 0.200 kg, h = 2.00 m and it is observed that the mass takes 1.90 s to reach the floor, how much energy was lost by friction?

Answer 1

a)

force equation for the mass hanging is given as

mg - T = ma

T = mg - ma eq-1

torque equation is given as

= T R = I

T R = I a/R

T = I a/R²

T = (MR²) a/R²

T = Ma eq-2

using eq-1 and eq-2

Ma = mg - ma

a = mg/(M + m)

a = (0.2 x 9.8)/(0.44 + 0.2)

a = 3.1 m/s²

using eq-2

T = M a

T = 0.44 x 3.1 = 1.4 N

b)

using conservation of energy

Potential energy = kinetic energy of hanging mass + rotational kinetic energy of drum

mgh = (0.5) m v² + (0.5) I w²

mgh = (0.5) m v² + (0.5) (MR²) (v/R)²

mgh = (0.5) m v² + (0.5) (M v²)

mgh = (0.5) (m + M) v²

(0.2) (9.8) (2) = (0.5) (0.2 + 0.44) v²

v = 3.5 m/s