A solid, uniform disk of radius 5.00 cm and mass 1.50 kg is rolling without slip

Question

A solid, uniform disk of radius 5.00 cm and mass 1.50 kg is rolling without slipping along a horizontal surface. The disk makes 2.00 revolutions per second. See Figure 3. Find the total kinetic energy (translational + rotational) of the disk. (The rotational inertia of a uniform disk of mass M and radius R about an axis passing through the center of the disk and perpendicular to the plane of the disk is given by I = 1/2 MR^2.) Find the minimum height h of the step that will prevent the disk from rolling past it. (Assume that the height h is adjusted so that the disk rolls just up to the top of the step and stops. Conserve energy.)

Explanation / Answer

N= 2 rev/s

so angular velocity= w= 2 pi N = 2 (3.14)(2) = 12.56 rad/s

radius of disc=5cm= 5/100 =0.05m

so velocity V = r w = 0.05 (12.56) = 0.628 m/s

total kinetic energy = 1/2 Iw2 + 1/2mv2

KEt= 1/2(1/2mR2)w2 +1/2mv2

now   V= Rw

so KEt= 1/4mv2 +1/2mv2

KEt= 3/4mV2

KEt= 3/4 (1.5)(0.628)2 = 0.44 joules

(b) using conservation of energy

KEt= mgh

0.44= (1.5)(9.8)h

h=0.03 m

or height = h= 3cm

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