A uniform solid sphere with mass M-2 kg rolls without slipping on horizontal sur

Question

A uniform solid sphere with mass M-2 kg rolls without slipping on horizontal surface so that its center of mass proceeds to the right with a constant linear a speed of 8 m/s. The moment of inertia of the uniform solid sphere is I_CM = 2 M R^2/5 where M is the mass and R is the radius of the uniform solid sphere and CM is center of mass. (a) The rotational kinetic energy of the uniform solid sphere is ________ (b) The total kinetic energy (rotation and translation) of the uniform solid sphere is ____ A uniform rod with weight 40 N is hung from the ceiling by a hinge at the upper end. One applies a horizontal force to the free end to hold the rod at an angle 30 degree with respect to the vertical. (a) The magnitude of this horizontal force is _____ (b) The magnitude of the force that the hinge exerts against the rod when the rod is held at this angle is ______ A simple pendulum with mass, m = 1 kg and length, L = 2.0 m is released by a push when the support string is at an angle of 25 degree from the vertical direction. The initial speed of the suspended mass at the release point is 3.0 m/s. (a) The maximum angle that pendulum moves in the second half of its swing is _____ (b) The mechanical energy of pendulum (measured relative to its lowest point) is ______

Explanation / Answer

problem 8

(a)

rotational kinetic energy Kr = (1/2)*I*w^2

Kr = (1/2)*(2/5)*M*R^2*(v/R)^2

Kr = (1/5)*M*v^2

Kr = (1/5)*2*8^2 = 25.6 J

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(b)

translational kinetic energy Kt = (1/2)*M*v^2

total kinetic energy Ktot = Kr + Kt = (1/5)*M*v^2 + (1/2)*M*V^2

Ktot = (7/10)*M*v^2 = (7/10)*2*8^2 = 89.6 J

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problem 9

along vertical

T*cos30 = W

along horizontal

T*sin30 = F

tan30 = F/W

F = W*tan30 = 40*tan30 = 23.1 N

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(b)

Tension T = W/cos30 = 40/cos30 = 46.2 N

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problem 10)

initial energy energy Ei = m*g*L*(1-costheta1) + (1/2)*m*v^2

final energy Ef = m*g*L*(1-costheta2)

Ef = Ei

m*g*L*(1-costheta2) = m*g*L*(1-costheta1) + (1/2)*m*v^2

1*9.8*2*(1-costheta2) = 1*9.8*2*(1-cos25) + ((1/2)*1*3^2)

theta 2= 47.4 degrees <<<<--------answer

(b)

at the lowest point

energy = m*g*L*(1-costheta1) + (1/2)*m*v^2 = 6.34 j

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