Question

1. (P) The figure below shows an 18.0kg crate of salami sliding down a ramp on a (massless) rope that is wrapped around a disk of radius 0.25m and mass 6.0kg. The coefficient of kinetic friction between the crate and ramp is μ k-0.24. What is the magnitude of the disk's angular acceleration? 30° A 3.1 rad/s2 B 6.7 rad/s2 C 9.8 rad/s2 D 11.4 rad/s2 E 13.7 rad/s2

Answer 1

Free Body Diagram Crate Disk 29 ay T mg Sine T Mg mgCose

Gravitational acceleration = g = 9.81 m/s²

Mass of the disk = M = 6 kg

Radius of the disk = R = 0.25 m

Moment of inertia of the disk = I

I = MR²/2

I = (6)(0.25)²/2

I = 0.1875 kg.m²

Mass of the crate = m = 18 kg

Angle of the incline = $\theta$ = 30^o

Normal force on the crate from the incline = N

Coefficient of kinetic friction between the crate and the incline = $\mu$_k = 0.24

Friction force on the crate = f = $\mu$_kN

Tension in the rope = T

Angular acceleration of the crate = $\alpha$

Acceleration of the block = a

a = $\alpha$R

From the free body diagram of the crate,

N = mgCos$\theta$

f = $\mu$_kmgCos$\theta$

ma = mgSin$\theta$ - f - T

ma = mgSin$\theta$ - $\mu$_kmgCos$\theta$ - T

T = mgSin$\theta$ - $\mu$_kmgCos$\theta$ - ma

For the disk,

I$\alpha$ = TR

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

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

(I + mR²)$\alpha$ = mgR(Sin$\theta$ - $\mu$_kCos$\theta$)

[0.1875 + (18)(0.25)²]$\alpha$ = (18)(9.81)(0.25)[Sin(30) - (0.24)Cos(30)]

$\alpha$ = 9.8 rad/s²

Magnitude of angular acceleration of the disk = 9.8 rad/s²