Question

1. A disk with a radius of 0.8 meters, a mass of 5 kg and c value of 1/2 is rolling without slipping at down an incline with a height (not length) of 6 meters. At the top of the incline, it is spinning at 16 rad/s. How fast is it moving in m/s (center of mass moving) at the bottom of the incline? Hint: first find how fast it's moving at the top of the incline as was done in the previous question.

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2. A uniform board with a length of 6 meters is supported horizontally by two supports. Support 1 is 0.5 meters from the left end and support 2 is 1.5 meters from the right end of the board. The normal force produced by support 1 is 53 Newtons. What is the weight of the board in Newtons?

Answer 1

1)
given
wi = 16 rad/s
R = 0.8 m/s

vi = R*wi

= 0.8*16

= 12.8 m/s

Apply conservation of energy

(1/2)*m*vf^2 + (1/2)*I*wf^2 = m*g*h + (1/2)*m*vi^2 + (1/2)*I*wi^2

(1/2)*m*vf^2 + (1/2)*(1/2)*m*R^2*wf^2 = m*g*h + (1/2)*m*vi^2 + (1/2)*(1/2)*m*R^2*wi^2

(1/2)*m*vf^2 + (1/4)*m*(R*wf)^2 = m*g*h + (1/2)*m*vi^2 + (1/4)*m*(R*wi)^2

(1/2)*m*vf^2 + (1/4)*m*vf^2 = m*g*h + (1/2)*m*vi^2 + (1/4)*m*vf^2

(3/4)*m*vf^2 = m*g*h + (3/4)*m*vi^2

(3/4)*vf^2 = g*h + (3/4)*vi^2

vf^2 = 4*g*h/3 + vi^2

vf = sqrt(4*g*h/3 + vi^2)

= sqrt(4*9.8*6/3 + 12.8^2)

= 15.6 m/s

2) let FL = 53 N

let FR is the force exerted by the right support.

Apply net torque about its center = 0

FL*(3 - 0.5) - FR*(3 - 1.5) = 0

FR = FL*2.5/1.5

= 53*2.5/1.5

= 88.3 N

so, Weight of the bar = FL + FR

= 53 + 88.3

= 141.3 N