Question

Problem 2; The 15.0 kg mass on a 40.0° frictionless inclined plane is tied to a string that is wrapped around a frictionless pulley (disk, I = MR2) of mass 12.0 kg and radius 9.00 cm. What is the angular acceleration of the disk after the 15.0 kg mass is released?

Answer 1

Free Bedy Diagram Blocks ac Polley T Smgsine T mg Cose Mg

Gravitational acceleration = g = 9.81 m/s²

Mass of the pulley = M = 12 kg

Radius of the pulley = R = 9 cm = 0.09 m

Moment of inertia of the pulley = I

I = MR²/2

I = (12)(0.09)²/2

I = 0.0486 kg.m²

Mass of the block = m = 15 kg

Angle of incline = $\theta$ = 40^o

Tension in the rope = T

Acceleration of the block = a

Angular acceleration of the pulley = $\alpha$

a = $\alpha$R

From the free body diagram of the block,

ma = mgSin$\theta$ - T

T = mgSin$\theta$ - ma

For the pulley,

I$\alpha$ = TR

I$\alpha$ = (mgSin$\theta$ - ma)R

I$\alpha$ = mgRSin$\theta$ - m($\alpha$R)R

(I + mR²)$\alpha$ = mgRSin$\theta$

[0.0486 + (15)(0.09)²]$\alpha$ = (15)(9.81)(0.09)Sin(40)

$\alpha$ = 50.04 rad/s²

Angular acceleration of the pulley = 50.04 rad/s²