Question

A pulley of moment of inertia 2.5 kg m2 is mounted on a wall as shown in the following figure. Light strings are wrapped around two circumferences of the pulley and weights are attached. Assume the following data: r, 49 cm, r, = 20 cm, m, = 1.0 kg, and m, = 2.5 kg. m2 mt (a) What is the angular acceleration of the pulley? (Enter the magnitude in rad/s2.) x rad/s2 0.0392 (b) What is the linear acceleration (in m/s2) of the weights? (Indicate the direction with the signs of your answers. Assume the upward direction is positive.) x m/s2 a,=0.019208 x m/s2 -0.00784 a

IM having a lot of trouble drawing out what is actually happening, and what up and down is in the picture.

Answer 1

here the pulley rotates clockwise.

given
I = 2.5 kg.m^2
r1 = 49 cm
m1 = 1 kg
r2 = 20 cm
m2 = 2.5 kg

a) let T1 and T2 are tensions in the strings

let a1 is the acceleration of block 1 and a2 is the acceleration of block 2

alfa is the angular acceleration of the block.

Net force acting on m1, Fnet1 = T1 - m1*g

m1*a1 = T1 - m1*g

T1 = m1*g + m1*a1 ------(1)

net force acting on m2, Fnet2 = m2*g - T2

m2*a2 = m2*g - T2

T2 = m2*g2 - m2*a2 ------(2)

net torque acting on pulley, Tnet = I*alfa

T2*r2 - T1*r1 = I*alfa

(m2*g2 - m2*a2)*r2 - (m1*g + m1*a1)*r1 = I*alfa

m2*g*r2 - m2*r2*alfa*r2 - m1*g*r1 - m1*r1*alfa*r1 = I*alfa

m2*g*r2 - m1*g*r1 = alfa*(I + m1*r1^2 + m2*r2^2)

==> alfa = (m2*g*r2 - m1*g*r1)/(I + m1*r1^2 + m2*r2^2)

= (2.5*9.8*0.2 - 1*9.8*0.49)/(2.5 + 1*0.49^2 + 2.5*0.2^2)

= 0.0345 rad/s^2 <<<<<<<<<<-----------------------Answer

b) a1 = alfa*r1

= 0.0345*0.49

= 0.0169 m/s^2 <<<<<<<<<<-----------------------Answer

a2 = -alfa*r2

= -0.0345*0.2

= -0.00690 m/s^2   <<<<<<<<<<-----------------------Answer