2 problems on simple harmonic motion

In summary: The axis of rotation is through the point of contact, so Angular acceleration is (K/m) and Linear acceleration is (K/m).
  • #1

Wen

44
0
these are 2 qn on simple harmonic motion which i am not sure if my solution is right or not.
1) A thin rod, length L hinged at one end, connected to a spring at the other end. Take the spring constant to be K. What is the angular frequency?

My solution:

let the displacement from equil. to be X
F=-kX
Torque= K.X.L=I a
a=K.X.L/I =L @ @ is the angul. aceleeration
@= KX/( 1/3 ML^2) = W^2X
W= (3k/ML^2)^0.5
2) Find the angular frequency of a small cylinder mass m, radius r, rolling in a cylindrical shell of radius R.

My sol.

aceleration=gsin@ let @ be the small change in angle
gsin@=R( Aceleration ,A)
A= gsin@/R
A= R angular acele. Since @is small, Sin@=@

g@/R=R angular a
angular a = w^2 @
w=(g/R^2)^0.5
 
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  • #2
1) angular frequency omega (w) should have units of [radian/sec].
Torque causes rotational inertia (I) to have angular acceleration alpha (@).
So you wanted to write kx L = I @ ... but this is rotation around the pivot.
The connection at the spring has acceleration a = L @ ,
and it is this acceleration that = - w^2 x ... [from 2nd derivitive of x(t)].

2) You never used the cylinder mass or radius, or its Inertia term "1/2". hmm
It might as well be sliding ... a = g sin(theta) is only for sliding or hanging.
at theta from the bottom, torque around contact point mg r sin(theta) = I alpha.
This alpha is the cylinder's rotational *spin* acceleration; "cylinder rolls" if
arclength s = R*theta = r*(phi+theta), where phi is the cylinder spin angle.
 
  • #3
For qn 2, if mgsin (theta)=I @
i can obtain the rotational angular acele
using mgsin phi= ma
i can obtain the translational; acele
but how should i equate to w^2x because are in different direction?
 
  • #4
First question 1

What about the mass and weight of the rod and its torque?
If the mass of the rod is zero then what will be its moment of inertia?

You have to consider rotetional SHM, similar equation with
torque proportional to - angular displacement
and with that you will get
freq. n = (1/2Pi) (K/I)^(1/2) Where K is torque constant.
 
  • #5
for qns 1 , i tried using 3 method,
1) KX= ma 3) KX=Ia
2) KX.L=I a 4) KX.L= ma

which one is wrong and why.
For qn 1, the spring is above the rod. Hence the rod is hanging from the spring.

Mukundpa, what is SHM?

Could someone enlighten me on qn 2?
 
  • #6
Wen said:
1) A thin rod, length L hinged at one end, connected to a spring at the other end. Take the spring constant to be K. What is the angular frequency?
Now you are saying "the spring is above the rod. Hence the rod is hanging from the spring."
So where is the hinge? Can you draw a diagram to show the position of spring and the rod?
 
  • #7
for qn 1, its a horizontal beam , pivoted at one end on the wall. Its other end is supported by a spring from the top so that it is horizontal.
 
  • #8
whether the other end is displaced slightly horizontally or vertically?
 
  • #9
If the end is displaced slightly such that the rod is rotated by small angle@ then the extension in the spring will be L@ and hence the tension in the spring is KL@.
This provides a restoring torque KL^2@ on the rod about the hinge.
So we can write the equation torque = - KL^2@ which is similar to F = - Kx

We can not take force Kx because there is a reaction force at the hinge which is not accounted.
 
  • #10
As for SHM angular frequency is given by squar rt (K/m) similarly for rotetional SHM it is given by squar rt (K/I) where K is torque constant and I is moment of inertia.

In your problem Torque constant is KL^2 and I = mL^2/3
 
  • #11
thanks. what about qn 2?

if i use gsin(thetre)= R angular a

since the cylinder also rotates abt the axis through its centre,

mgsin( thetre).r=torque=I. angular acele abt its axis.
As the two acele. are in different directions, how should i combine them together?
 
  • #12
Rolling without slipping is not possible without friction. It will be better to find acceleration of centre of mass of the sphere. Do you have any idea for that?


Consider the axis of rotation passing through the point of contact and find angular acceleration and then linear acceleration.
 
  • #13
okay, if i find the angular aceleration and translational a, should i sum them up to equal to w^2 thetre. But the direction of the acele is different.
 
  • #14
Let the line joining the axis of the small cylinder and the axis of the surface makes an angle @ with vertical. the torque on the small cylinder about the point of contact is
mgr sin@
and angular acceleration is
mgrsin@/(3mr^2/2) = 2gsin@/3r
hence linear acceleration of center of mass will be...( radius R-r)
2gsin@/3 = 2g@/3 = 2g[x/(R-r)]/3
where x is displacement of C.M. from equilibrium position and as @ is small enough
Hence the restoring force F = - m[2g/3(R-r)] x which is the equation of SHM
so angular frequency is sq.rt.of 2g/3(R-r).

EEEnjoy !
 
  • #15
isn't the I of cylinder 1/2mr^2.

However, if i use the same way to work out one qn ( which is very popular in all textbooks) A solid sphere radius R rolls without slipping in a cylinderical trough, radius 5R. show that the period is T=2pi (28R/5g)^0.5.

Since Rmgsin@=I a.
a= Rmgsin@/(2/5 mR^2)=5gsin@/2R=5g@/2R
@= X/(5R-R)
a= 5g/2R(X/4R)=w^2X

But cannot prove to the eqn stated.
 
  • #16
Torque is taken about the point of contact hence the I is to be taken about the same point i.e. (according to parallel axes theoren) 1/2mr^2 + mr^2 = 3/2 mr^2
 
  • #17
Wen said:
A solid sphere radius R rolls without slipping in a cylinderical trough, radius 5R. show that the period is T=2pi (28R/5g)^0.5.

Since Rmgsin@=I a.
a= Rmgsin@/(2/5 mR^2)=5gsin@/2R=5g@/2R
@= X/(5R-R)
a= 5g/2R(X/4R)=w^2X

But cannot prove to the eqn stated.

The simelar problem
The moment of inertia of the sphere about a tangent is

[tex]\frac{\ 2mR^2}{5} + mR^2 = \frac{\ 7mR^2}{5}[/tex]
 
Last edited:
  • #18
thank you . thank you
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion where an object oscillates back and forth around a central equilibrium point due to the restoring force of a spring or pendulum.

2. What are the equations for simple harmonic motion?

The equation for displacement of an object undergoing simple harmonic motion is x(t) = A*cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle. The equation for the velocity of the object is v(t) = -A*ω*sin(ωt + φ), and the equation for the acceleration is a(t) = -A*ω^2*cos(ωt + φ).

3. How is simple harmonic motion related to energy?

Simple harmonic motion involves the exchange of potential and kinetic energy as the object oscillates. At the equilibrium point, the object has maximum potential energy and zero kinetic energy. As it moves away from the equilibrium point, potential energy decreases and kinetic energy increases. At the maximum displacement, the object has maximum kinetic energy and zero potential energy. The total energy remains constant throughout the motion.

4. What are some real-life examples of simple harmonic motion?

Some examples of simple harmonic motion in everyday life include a swinging pendulum, a mass-spring system, and a vibrating guitar string. Other examples include the motion of a grandfather clock pendulum, the motion of a bungee jumper, and the motion of a car's suspension system.

5. How does amplitude affect simple harmonic motion?

The amplitude of simple harmonic motion is the maximum displacement from the equilibrium point. A larger amplitude results in a longer period and a smaller amplitude results in a shorter period. The amplitude also affects the maximum potential and kinetic energies of the system.

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