Hey guys sorry kind of long but I am completely lost any help would be much appr
ID: 1314812 • Letter: H
Question
Hey guys sorry kind of long but I am completely lost any help would be much appreciated and will recieve stars!!
(1) An object starts from rest and travels around a circle of radius 10 cm with constant angular acceleration. The object covers an angular distance of 2 rad in 16 s. What is the tangential speed of the object after it has moved the 2 rad?
(A) 2.5 cm/s (B) 3.0 cm/s (C) 3.5 cm/s (D) 4.0 cm/s (E) 4.5 cm/s
(2) The rotational kinetic energy of a uniformly dense solid sphere with a mass of 2 kg and a radius of 20 cm rotating around an axis through its center is given by the following equation
Krot(t)= (5J/s^2)t^2
How many revolutions does the sphere complete between t = 2 s and t = 4 s?
(A) 15.3 rev (B) 16.0 rev (C) 16.9 rev (D) 18.0 rev (E) 19.9 rev
(3) A uniformly dense solid sphere is rolling without slipping down a ramp around an axis through its center. As the sphere rolls gravitational potential energy is being converted into translational kinetic energy and rotational kinetic energy. What fraction of the gravitational potential energy is converted into rotational kinetic energy?
(A) 2/7 (B) 1/3 (C) 2/5 (D) 1/2 (E) 3/5
(6) An object oscillates in one-dimension with an amplitude of 8 cm. At t = 0 s, its position is x = 4 cm and it is moving with a velocity of -6 cm/s. What is the velocity of the object at t = 3 s?
(A) 1.8 cm/s (B) 2.2 cm/s (C) 2.6 cm/s (D) 3.0 cm/s (E) 3.4 cm/s
(7) The energy of an isolated system oscillating along the x-axis is given by the following equation
E=(4kg)vx2+(10J/m2)x2
What is the period of the oscillation?
(A) 3.5 s (B) 3.9 s (C) 4.3 s (D) 4.7 s (E) 5.1 s
(8) A simple pendulum is constructed from a 2 kg bob and a 1.5 m long massless string. How long will it take this system to complete 3 small amplitude oscillations?
(A) 7.4 s (B) 8.0 s (C) 8.6 s (D) 9.2 s (E) 10.0 s
(9) A 7.5 kg block is attached to a horizontally-mounted spring with a spring constant of 750 J/m2. The block is released from rest with the spring extended by 15 cm from its normal length of 27 cm. The horizontal surface along which the block moves is frictionless. What is the maximum speed of the block?
(A) 1.0 m/s (B) 1.5 m/s (C) 2.0 m/s (D) 2.5 m/s (E) 3.0 m/s
(10) A block with a mass of 4 kg is being pushed up a 30
Explanation / Answer
1) so theta = w0 t + 1/2 a t^2
we know 20 = 0
so theta = 1/2 a t^2
2 = 1/2 a 16^2
a=1/64
w = w0 + a t = 1/64*16 = 1/4 rad/s
v = w r = 0.25*10 = 2.5 cm/s
so a)
2)
K rotation = 1/2 I w^2 = 5 t^2
I = 2/5 M r^2 = 2/5*2*.2^2= 0.032
0.5*0.032*w^2 = 5 t^2
w = sqrt(5/(0.5*0.032)) t = 17.68 t
so alpha = 17.68
theta(t=4) = 0.5*17.68*4^2
theta(t=2) 0.5*17.68*2^2
dtheta = 0.5*17.68*(4^2-2^2)= 106 rad = 16.87 rev so C)
3)
so Ei = Ef
m gh = 1/2 m v^2 + 1/2 I w^2
but I = 2/5 M R^2
and w = v/r
1/2 I w^2 = 1/2 2/5 M R^2 (v/R)^2 = 1/5 M R^2
so fraction of rotation is
1/5/(1/5 + 1/2)= 1/5/( 7/10) = 2/7 so A)
6)
x = A cos(wt + phi)
at t = 0, 4 = 8 cos(phi)
phi = 1.047
v = dx/dt = - A w sin(wt + phi)
so at t = 0
v = -8*w sin(1.047)=-6
w = 0.866
v(t = 3) = -8*0.866*sin(0.866*3+1.047)= 3.4 cm/s
so E)
7)
so E = 1/2 mv^2 + 1/2 k x^2
so we can see that k = 20 and m = 8
T = 2 pi sqrt( m/k) = 2*pi*sqrt(8/20)= 3.9 so B)
8) t = 2*pi*sqrt(L/g) = 2*pi*sqrt(1.5/9.81)= 2.46
we want three times that so A)
9)
Ei = Ef
1/2 k A^2 = 1/2 mv^2
0.5*750*.15^2 = 0.5*7.5*v^2
v= 1.5 m/s
so B)
10)
Ei + Wforce+ W friction = Ef
0 + F*7 - 8*7 = 1/2 mv^2 + m g h
F*7 - 8*7 = 0.5*4*4^2 + 4*9.81*7*sin(30 degrees)
F=32.2 N so E)