Question

I have solved A AND B. I need the equation to set up section C. I know that the moment of inertia for a cylinder is 0.5mr^2 PLEASE HELP

A light string is wrapped around a solid cylinder and a 300 g mass hangs from the free end of the string, as shown. When released, the mass falls a distance 54 cm in 3.0 s.

a.) Draw free-body diagrams for the block and the cylinder.

b.) Calculate the tension in the string.

c.) Calculate the mass of the cylinder.

Answer 1

C)
let
m = 300 g = 0.3 kg

d = 54 cm

t = 3 s

let a is the acceleration of the hanging block

let M is the mass of cyllinder and r is the radius of the cyllinder.

use, d = u*t + (1/2)*a*t^2

d = 0 + (1/2)*a*t^2

==> a = 2*d/t^2

= 2*0.54/3^2

= 0.12 m/s^2

Let T is the tension in the string.

net force acting o hanging block, Fnet = m*g - T

m*a = m*g - T

==> T = m*g - m*a

net torque acting on cyllinder, Tnet = I*alfa

T*r = I*a/r

(m*g - m*a)*r = (1/2)*M*r^2*a/r (since a_tan = r*alfa)

m*g - m*a = (1/2)*M*a

M = 2*(m*g - m*a)/a

= 2*(0.3*9.8 - 0.3*0.12)/0.12

= 48.4 kg <<<<<<<<<<<<<<--------------------------Answer