# Rigid Body Dynamics

(Please refer attachment for detailed problem description)

Problem 1 : The 40 kg disc is released from rest with the spring compressed. At the instant shown (see attachment) it has a speed of 4 m/s and the spring is unstretched. From this poinr determine the distance d that the disc moves down the 30 degree ramp before momentarily stopping. The disc rolls without slipping. Determine :

a) The distance d in meters down the ramp from point A.

b) The moment of inertia of the disc about G in kgm^2

c) The angular velocity of the disc in rad/sec at this instant.

d) The work done by the spring in Nm during the downward motion.

e) The distance h in m, up the ramp from A to the starting point.

Problem 2 : A 5 kg uniform rod 800mm long rotates about pin B in a vertical plane. If the rod is released from rest when it is horizontal, determine :

a) Moment of inertia of the rod about B in kg m^2.

b) Angular velocity of the rod in rad/sec when vertical.

c) Angular velocity of the rod in kg m^2 at theta = 75 degrees

d) X direction support reaction in N at pin B at theta = 75 degrees

e) Y direction support reaction in N at pin B at theta = 75 degrees

https://brainmass.com/math/discrete-math/rigid-body-dynamics-175831

#### Solution Preview

Please refer attachment.

Problem 1

k = 200 N/m

4 m/s

G Mass = 40 kg, Radius = 0.3m

A d

dsin30O

30O

Moment of inertia of the disc about an axis through centre of the disc = I = ½ MR2 = ½ x 40 x 0.32 = 1.8 kgm2

At the given moment the disc has a translational velocity v = 4 m/s.

Its angular velocity ω = v/R = 4/0.3 = 13.33 radians/sec

At the given moment the disc has a translational kinetic energy = ½ mv2 = ½ x 40 x 42 = 320 J

And rotational KE = ½ Iω2 = ½ x 1.8 x (13.33)2 = 160 J

Total KE = 320 + 160 = 480 J

With reference to the centre of the disc at the lowest position, the height of the disc in the given position is dsin30O = 0.5d.

Potential energy of the disc in the given position with reference to its lowest position = mgh = 40x9.81x0.5d = 196d

Total mechanical energy of the disc in the given position = KE +PE ...

#### Solution Summary

A rigid body dynamics is analyzed. The moment of inertia of the disc about G in kgm^2 is analyzed.