Download HW#10b Note: numbers used in solution steps are different from

HW#10b
Note: numbers used in solution steps are different from your WebAssign values.
Page 1 of 6
a). For Spring 2, Spring force: F=k*∆x, ∆x=0.6-0.1 =0.5 (m), Spring Force = object weight = 3 (N)
k = F/ ∆x =3/0.5= 6 (N/m)
b.) For Spring 3, F= k*∆x, ∆x=1.1-0.1 =1.0 (m),
Object weight= Spring Force= k*∆x= 6* 1=6 (N)
In orbit, he has centripetal acceleration, so the total net force along the radius direction pointing to the earth
center is NOT zero. Hence if he stands on a scale, the normal force from the scale is not equal to mg. Also when
he is in orbit, g is NOT equal to 9.8 m/s2
a). A mass on spring: Oscillation period:
, So:
2
2
=3000*1 / (4*3.14 )=76 kg
So:
b). Potential energy stored in the spring U =1/2* k*∆x 2
When it is Furthest from the equilibrium length: ∆x=Amplitude=4m
U =1/2* k*Α 2=0.5*3000*42=24000 J
c). When it passes equilibrium: All Potential energy is converted into Kinetic energy:
½*m*V2=Umax = 1/2* k*Α 2=24000 J
V2=k*A2/m
d). oscillation period is in dependant to amplitude. It is only determined by the stiffness of the spring (spring
constant) and the mass of the object. So T is still 1 s.
HW#10b
Note: numbers used in solution steps are different from your WebAssign values.
Page 2 of 6
Problem 3. The height of a tower is measured by attaching a simple pendulum to its
ceiling, whose length is barely enough to stay off the floor. The pendulum is let go
from a small angle, and takes 12 s to return to the same location it started from.
m
a.) How tall is the tower?
If the pendulum mass is let go 0.2 m above the floor,
b.)how fast is the mass traveling as it grazes the floor?
m/s
If the the pendulum mass is let go 0.3 m above the floor, what would be the period
to finish a cycle?
Solution. (a) The period of the pendulum is related to the length of the
pendulum, H=L in this case, by
L
g
h
.
So, the height must be
(b) This is a matter of conservation of energy. The potential energy when it is released is converted
into kinetic energy at the bottom of its motion. Ei=Ef. Hence, PEmax at the release point is equal to
KEmax at the bottom point.
mgh = 12 mv 2
giving
(
)
v = 2 gh = 2 9.8 m/s 2 ( 0.2 m ) = 1.980 m/s
For question 3b, think about your object. There is no friction. Wnc=0. Energy converts between PE and KE
again and again. At the beginning when it is let go, the total energy PE+KE was only PE=mghmax, later when it
is at bottom the total Energy was only KE=1/2 mvmax2
Here all PE were converted to KE. v and KE reach maximum and PE reaches zero.
Notice that, this h, the maximum height of the object above the bottom. It determines the amplitude but it
doesn’t affect the period T. If you change h, by releasing it at 0.3 m above the ground and release, the speed at
the bottom will increase. But the period is still
Period of pendulum is only determined by the total length of the string and g, independent to
the initial height, as long as it is small angle.
Question 2c is very similar. In both 3b and 2c, you should notice that total KE+PE do not change. How much
you reduce in PE is equal to how much you increase in KE. Also, PE and KE are all scalars. We do not worry
about their direction and "components".
Notice the special points at equilibrium point, PE=0 KE=totalE and reaches maximum.
And at locations farthest from equilibrium, KE=0 and PE=total E.
At those locations, sin (ω*t) or cos(ω*t) are either zero or 1. No need to worry about sin or cos for them.
HW#10b
Note: numbers used in solution steps are different from your WebAssign values.
Page 3 of 6
Problem 4.
Problem 5.
A is the amplitude
One period of time, it moves through distance of 4 times A.
0.5 T, it moves through a distance of twice of A.
0.25 T, it moves for one A
For 38 A, 38/4= 9.5 T. 9 and a half period is the time needed to move 38 A.
Problem 6.
Visualize the motion, within 1 period of time, the object covers total distance of 4A. d=4A
(not that 1T=4A, Time is never equals to length). Within 2 T of time the object covers 8A of distance.
Within 5/2 T, the object covers 2.5*4A=10A distance.
If that distance is 14D, you know that 10A=14D, and one A =1.4D
Problem 7.
HW#10b
Note: numbers used in solution steps are different from your WebAssign values.
Page 4 of 6
At x=0, no potential energy, all energy is kinetic energy. V= Vmax
At x=0, no resorting force, also no acceleration, a=0
At x=A, no kinetic energy, all energy is potential energy. V=0
At x=A, resorting force = - k A, reaches it’s maximum magnitude, and along the negative direction. ( opposite
direction to A).
So, a= -1 * amax
The Following conceptual questions are very important for quiz and exams.
Problem 8
Key: The period of Simple harmonic motion is independent to amplitude. When Amplitude is larger, the initial
pertential energy ½ k A2 saved in the spring or mgh saved in pendulum is larger, hence the kinetic energy the
object gets at any given time is also larger. Hence the speed is faster. Even though the distance traveled by the
object within one period is doubled, its speed at any given time is also doubled, and it finish one cycle at the
same T, independent of initial position or Amplitude.
HW#10b
Note: numbers used in solution steps are different from your WebAssign values.
Page 5 of 6
On this page, there is One extra problem for those who are interested. Not required.
a). Potential energy stored in the spring U =1/2* k*∆x 2
Initially, ∆x = 0.7 m, U =1/2* k*∆x 2 = 0.5*50*0.72 = 12.2 ( J )
When the pinball is fired all these energy is converted to the pinball’s kinetic energy.
b). When it rolls, it doesn’t slip, at the maximum height the mass center of the ball’s velocity is zero, it doesn’t
climb anymore, also it stops rotating. Because the mass center speed v = ω* r.
When v = 0, ω=0, no kinetic energy is left. Non-slip rolling doesn’t lose total mechanical energy because no
non-conservative force does any work to steal or increase its energy. All energy is converted to gravitational
potential energy at this point. (The friction at the contact point is static friction, because the contacting point has
no motion during non-slip rolling. hence friction doesn’t do work. ).
So, at the maximum height, mgh = initial U =1/2* k*∆x 2 =12.2 (J)
So, h= 12.2/mg= 12.2/(0.2*9.8) = 6.25 ( m)
c). When the Pinball is 2.5 meter above its initial position, part of initial energy was converted to potential
energy mgh and the remaining was still kinetic energy.
HW#10b
Note: numbers used in solution steps are different from your WebAssign values.
From bottom to 2.5 m, Ei=Ef =12.2 J= Kf+Uf, so,
Kf = Ef - Uf = 12.2 – m g h = 12.2 – 0.2*9.8*2.5 = 7.3 (J)
Also Kf = ½ m Vcm2 + ½ I ω2
While, I = (2/5) m r2 ;
So: Kf = ½ m Vcm2 + ½ I ω2 = ½ m Vcm2
;
ω = Vcm/r ;
+ ½ * (2/5) * m r2 ω2 = (7/10)*m* Vcm2
Vcm2 =(10/7)*Kf /m = (10/7)*7.3/0.2 =52.1 m2/s2
Vcm = 7.23 m/s
ω = Vcm/r = 7.23/0.3 = 24.1 rad/s
Vcm= ω * r
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