A thin, cylindrical rod script i = 26.6 cm long with a mass m = 1.20 kg has a ba

Question

A thin, cylindrical rod script i = 26.6 cm long with a mass m = 1.20 kg has a ball of diameter d = 10.00 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?
Correct: Your answer is correct. J

(b) What is the angular speed of the rod and ball?
Correct: Your answer is correct. 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 31.6 cm?
vswing is . vfall by ?%.

Explanation / Answer

this might help,i had a similar problem....


A thin, cylindrical rod 36.1 cm long has mass 1.5 kg and radius 1.4 cm. A 8.3 kg ball of diameter 10.8 cm is attached to one end. The arrangement is originally vertical with the ball at the top and is free to pivot about the other end.

a) After the ball-rod system falls a quarter turn, what is its rotational kinetic energy? The acceleration of gravity is 9.8 m/s^2 . Answer in units of J.

b) At the same point, what is its angular speed? Answer in units of rad/s.

c) At the same point, what is the linear speed of the ball? Answer in units of m/s.

Step by step calculations.







g = 9.8 m/s^2
m = mass of the rod = 1.5 kg
M = mass of the ball = 8.3 kg
L = length of the rod = 36.1 cm = 0.361 m
R = radius of the ball = 10.8/2 = 5.4 cm = 0.054 ,m
X = distance from pivot to center of mass = 37.91 cm = 0.3791 m (see Note)
w = angular velocity
a = linear acceleration of center of mass
V = velocity of center of mass = w*R
D = distance from pivot point to center of sphere = L + R = 41.5 cm = 0.415 m

Use conservation of energy.
The change in energy is the change in gravitational potential energy between the starting point and the current point at angle T. We are doing this for the center of mass of the system.

d(PE) = PE(at T) - PE(top)

PE(top) = (m + M)*g*X
PE(at T) = (n + M)*g*X*cos(T)
d(PE) = (m + M)*g*X[cos(T) - 1]

Conservation of energy now tells us that the change in potential energy must equal the change in kinetic energy.

d(KE) = (1/2)I*w^2

I is the total moment of inertia of the rod and ball which is just the sum of the moments of inertia for each. For the rod we can use the thin rod formula but the ball isn't exactly a point so I will have to integrate. I give the answer and an explanation but I can append the calculations if you want them.

Irod = (1/3)m*L^2 ................... using the thin rod equation
Isphere = M*[D^2 + 2R^2/5] .... not exactly a point so I had to do the integration (I finally found the general rule and it is "the moment of inertia for a rigid body about any axis equals moment of inertia of that body around its own center of mass, plus the moment of inertia of a point mass about the axis of rotation." So I did the integration correctly). If you want details I will be gald to pass them along.

I = Irod + Isphere = (1/3)m*L^2 + M*(D^2 + 2R^2/5)
d(KE) = (1/2)[(1/3)m*L^2 + M*(D^2 + 2R^2/5)]*w^2

Conservation of energy, equate the potential and kinetic energies.
d(PE) + d(KE) = 0
(m + M)*g*X[1 - cos(T)] = (1/2)[(1/3)m*L^2 + M*(D^2 + 2R^2/5)]*w^2

And solve for w^2 as a function of T.
w^2 = 2*(m + M)*g*X[1 - cos(T)] / [(1/3)m*L^2 + M*(D^2 + 2R^2/5)]

(B) Just substitute in the values and do some math. We want it after 1/4 turn so T = pi/2
w^2 = 48.40687265
w = 6.958 rad/s

(A) KE = (1/2)I*w^2
I = (1/3)m*L^2 + M*(D^2 + 2R^2/5) = 1.504
KE = (1/2)*1.504*(6.9575)^2 =36.41 J

This should also be the change in potential energy.
PE = (m + M)*g*X = (1.5 + 8.3)*9.81*1.504 = 36.41 J

(C) Linear speed = w*D = 6.9575*0.415 m/s
Linear speed = 2.887 m/s

Note: Center of mass.
m*L/2 + (L + R)M = X(m + M)
X = [m*L/2 + (L + R)M]/(m + M)
X = [1.5*(36.1/2) + (36.1 + 5.4)*8.3]/(1.5 + 8.3) = [1.5*18.05 + 41.5*8.3]/9.8 = 371.525/9.8
X = 37.91 cm

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