Massive Pulley: Consider two masses attached by an ideal massless rope going ove

Question

Massive Pulley: Consider two masses attached by an ideal massless rope going over a massive pulley as shown below where mass one is 1.5 kg, mass two is 8.3 kg and the pulley is 1.2 kg with a radius of 12 cm. with a frictionless incline at 54 degrees. What is the Net force on mass 1 ? What is the tension in the rope connected to mass 1? What is the linear acceleration of mass 2? What is the angular acceleration of the pulley? Integrate the acceleration equation to find the velocity at 2.3 s if the system started at rest.

Explanation / Answer

m1 = 1.5 kg
m2 = 8.3 kg
pulley mass, m = 1.5 kg, radius, r = 12 cm = 0.12 m

moment of inertia of pulley, I = 0.5*m*r^2 = 0.5*1.2*0.12^2 = 0.00864 kg.m^2

let T1 is the tension in vertical string and T2 is the tension in horizontal sting.

let a is the linear acceleration of the bodies.

net force acting on m1

Fnet = T1 - m1*g

m1*a = T1 - m1*g

==> T1 = m1*a + m1*g --(1)

net force acting on m2

Fnet = m2*g*sin(54) - T2

m2*a = m2*g*sin(54) - T2

==> T2 = m2*g*sin(54) - m2*a --(2)

angular acceleration of whell, alfa = a/r

net torque acting on pulley

net totorque = T2*r - T1*r

I*alfa = T2*r - T1*r

I*a/r = T2*r - T1*r

I *a = T2*r^2 - T1*r^2

I*a = (m2*g*sin(54) - m2*a)*r^2 - (m1*a + m1*g)*r^2

0.00864*a = (8.3*9.8*sin(54) - 8.3*a)0.12^2 - (1.5*a + 1.5*9.8)*0.12^2

0.00864*a = 0.948 - 0.1195*a - 0.0216*a - 0.21168

(0.00864 + 0.1195 + 0.21168)*a = 0.948 - 0.21168


a = 2.17 m/s^2


A) Fnet on m1 = m1*a = 1.5*2.17 = 3.255 N

B) T1 = m1*a + m1*g = 1.5*(2.17 + 9.8) = 11.96 N

C) a = 2.17 m/s^2

D) alfa = a/r = 2.17/0.12 = 18.08 rad/s^2


E) v = u + a*t

= 0 + 2.17*2.3

= 4.991 m/s

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