Displayed in the figure a uniform solid cylinder and a hollow hoop have the same
ID: 1623612 • Letter: D
Question
Displayed in the figure a uniform solid cylinder and a hollow hoop have the same mass and radius. Both initially at rest are released simultaneously from the same height on a ramp. Both are rolling down the ramp without slipping. Choose the correct statement with regard to this situation. a. The hollow cylinder has a higher total kinetic energy in the bottom of the ramp. b. The solid cylinder has a higher total kinetic energy in the bottom of the ramp. c. Both cylinders arrive the bottom of the ramp simultaneously. d. The hollow cylinder reaches the bottom of the ramp first. e. The solid cylinder reaches the bottom of the ramp first. The figure below displays a 90.0 kg person initially sprinting at a speed v_0 = 8.00 m/s then leaping tangentially onto the edge of a 250 kg merry-go-around of radius 2.00 m, which is originally at rest. Treating the merry-go-round as a uniform disk, and the person as a point mass on the rim, find the resulting angular velocity of the merry-go-around. a. 0.458 rad/s b. 1.67 rad/s c. 2.88 rad/s d. 1.54 rad/s e. 3.20 rad/s A uniform steel beam of length L = 60 m and mass m_1 = 120 kg is attached via a hinge to the side of a building. The beam is supported by a steel cable attached to the end of the beam at an angle theta = 30 degree, as shown in the figure. Through the hinge, the wall exerts an unknown force F, on the beam. A workman of mass m_2 = 75 kg sits eating lunch a distance d = 1.8 m form the building. Determine the tension exerted by the cable on the beam. a. 1913 N b. 1004 N c. 1521 N d. 1619NExplanation / Answer
15. from energy conservation, Gravitational PE will get converted into Kinetic energy.
and initial GPE = m g h { h is same for both }
hence both will have same kinetic energy.
but total kinetic energy = translational + rotational KE
= m v^2 /2 + I w^2 /2
= m v^2 /2 + (I / r^2) v^2 / 2
for hollow I/r^2 = m
for solid, I/r^2 = m/2
v will be greater for solid sphere.
v^2 - 0^2 = 2 a d
a = v^2 / 2 d
vf = vi + a t
v = 0 + (v^2 / 2d)t
t = 2 d / v
v is greater for solid sphere hence time will be less.
hence solid sphere will reach first.
Ans(E)
16. applying angular momentum conservation,
90 x 8 x 2 = [ (90 x 2^2) + (250 x 2^2 / 2)] w
w = 1.67 rad/s
Ans(b)
17. balancing moment about the hinge,
(1.8 x 75 x 9.8) + (3 x 120 x 9.8) - (6 T sin30) = 0
T = 1619 N
Ans(d)