Question

A uniform disk with radius 0.440 m and mass 32.0 kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to θ(t)=( 1.00 rad/s )t+( 8.10 rad/s2 )t2 .

What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.300 rev ?

Express your answer with the appropriate units.

Answer 1

A uniform disk with radius 0.440 m

mass = 32.0 kg

θ(t)=( 1.00 rad/s )t+( 8.10 rad/s2 )t²

the disk has turned through 0.300 rev

that is    θ(t) = 2*3.14*0,300 radian = 1.884 rad

Put this in the above equation we get

1.884 = t + 8.10 t²

Solving above quadratic equation we get

t = 0.4245 seconds

w = dθ/dt = 1 + 16.2t

the resultant linear acceleration of a point on the rim of the disk

resultant linear acceleration =( radial acceleration² + tangential acceleration² )^1/2

radial  Acceleration = w²R = (1 + 16.2t )² *  0.440

put t= 0.4245

radial Acceleration = 27.3 m/s2

d²θ/dt² = 16.2

tangential acceleration = 0.440(d²θ/dt²) = 7.128 m/s²

resultant linear acceleration = (7.128 ² + 27.3 ² )^1/2 = 28.22 m/s²