(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
Please refer attachment.
k = 200 N/m
G Mass = 40 kg, Radius = 0.3m
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 ...
A rigid body dynamics is analyzed. The moment of inertia of the disc about G in kgm^2 is analyzed.