A thin, cylindrical rod ? = 20.8 cm long with a mass m = 1.20 kg has a ball of d

Question

A thin, cylindrical rod ? = 20.8 cm long with a mass m = 1.20 kg has a ball of diameter d = 6.00 cm and mass M = 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top as shown in the figure below. The combination is free to pivot about the bottom end of the rod after being given a slight nudge.

(a) After the combination rotates through 90 degrees, what is its rotational kinetic energy?

________ j

(b) What is the angular speed of the rod and ball?

________ rad/s

(c) What is the linear speed of the center of mass of the ball?

_________ m/s

(d) How does it compare with the speed had the ball fallen freely through the same distance of 23.8 cm?

Vswing is greater than Vfall by _____ %

Explanation / Answer

here,

Length of rod, l = 20.8 cm = 0.208 m
mass of rod, Mr = 1.20 kg
radius of ball,r = 6/2 = 3cm = 0.03 m
mas of ball, Mb = 2.0 kg

Moment of inertia of system :
I = Irod + Iball
I = 1/3*Mr*l^2 + (2/5)Mb*R^2 + Mb(L + d/2)^2
I = 0.333*1.20*(0.208)^2 + 0.4*2*(0.03)^2 + 2(0.208 + 0.03 )^2
I = 0.1313 kg.m^2

Part A :
From Conservation of energy : ROtational KE = Change in PE
RKE = PE(rod) + PE(ball)
RKE = Mr*g*l/2 + Mb*g(l+r)
RKE = 1.20*9.8*0.208/2 + 2*9.8(0.208 + 0.03)
RKE = 5.887 J

PART B:
Rotational Kinetic Enrgy = 0.5 * I * w^2

solving for angular speed, w
w = sqrt(2*RKE/I)
w = sqrt(2*5.887/0.1313)
w = 9.47 rad/s

PART C:
as linear speed = w/D ( where D = L + r)
v = w*(l+r)
v = 9.47 *(0.208 + 0.03)
v = 2.254 m/s

PART D:
From conservation of erngy we have :
Ke gained by object = Potential energy lost by object
0.5 * m * v^2 = mgh

soving for velocity, v
v = sqrt(2*g*h)
v = sqrt( 2*98*(0.208+0.03) )
v = 6.83 m/s

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