- #1
Abhishekdas
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Rotational mechanics...
This is actually a simple harmonic motion question but the doubt i have is with a concept in rotation...
I will state the exact problem...
A solid sphere(radius R) rolls without slipping in a cylindrical trough(Radius 5R).
Find the time period of small oscillations...
refer to diagram in attachment if necessary...
Now as i have told you the shm part is not bothering me much but its the rotation part...
Anyway i will state what i did...
In my calculations
w1=angular vel of sphere about its own CM
w2= ......about the centre of the trough
alpha1=angular acceleration of sphere wrt its own CM
alpha2=......... the centre of the trough
v=vel of CM of sphere
now for pure rolling v=Rw1
and v also = 4R *w2 ...using v=rw wrt the centre of the trough
equating v from the two equations we get w1=4*w2
differentiating wrt time alpha1=4*alpha2
PART 1
Here i am dealing with the torque causing pure rolling on the surface of the trough...I have assumed that the sphere has been displaced from its mean position ie at the bottom of the trough by a small angle theta...
now... mgsin(theta) - f=ma.....f=friction m=mass of sphere...
f*R=I*alpha1 ...torque equation for rolling here I = 2MR2/5
solvin them we get f=2mgsin(theta)/7 and a=5gsin(theta)/7
PART 2
Here i am dealing with the torque which causes the sphere to revolve about the centre of the cylinder...
mgsin(theta)*4R - f*5R=I*alpha2 ...I= using paralle axes thoerem = 82MR2/5
sustituting value of f and simplifying we get
mgsin(theta)*18/7=(82/5)*MR2*alpha2
solving we get alpha2=45*gsin(theta)/287R
and alpha1 =a/R (as a=R*alpah1) ...alpha1 = 5gsin(theta)/7R...
now here is the problem...alpha1 is not coming out to be equal to (alpha2)*4... why is it so?
Please take the pain of reading this entire thing ...I would be really grateful if you help me out...thank you
Homework Statement
This is actually a simple harmonic motion question but the doubt i have is with a concept in rotation...
I will state the exact problem...
A solid sphere(radius R) rolls without slipping in a cylindrical trough(Radius 5R).
Find the time period of small oscillations...
refer to diagram in attachment if necessary...
Homework Equations
The Attempt at a Solution
Now as i have told you the shm part is not bothering me much but its the rotation part...
Anyway i will state what i did...
In my calculations
w1=angular vel of sphere about its own CM
w2= ......about the centre of the trough
alpha1=angular acceleration of sphere wrt its own CM
alpha2=......... the centre of the trough
v=vel of CM of sphere
now for pure rolling v=Rw1
and v also = 4R *w2 ...using v=rw wrt the centre of the trough
equating v from the two equations we get w1=4*w2
differentiating wrt time alpha1=4*alpha2
PART 1
Here i am dealing with the torque causing pure rolling on the surface of the trough...I have assumed that the sphere has been displaced from its mean position ie at the bottom of the trough by a small angle theta...
now... mgsin(theta) - f=ma.....f=friction m=mass of sphere...
f*R=I*alpha1 ...torque equation for rolling here I = 2MR2/5
solvin them we get f=2mgsin(theta)/7 and a=5gsin(theta)/7
PART 2
Here i am dealing with the torque which causes the sphere to revolve about the centre of the cylinder...
mgsin(theta)*4R - f*5R=I*alpha2 ...I= using paralle axes thoerem = 82MR2/5
sustituting value of f and simplifying we get
mgsin(theta)*18/7=(82/5)*MR2*alpha2
solving we get alpha2=45*gsin(theta)/287R
and alpha1 =a/R (as a=R*alpah1) ...alpha1 = 5gsin(theta)/7R...
now here is the problem...alpha1 is not coming out to be equal to (alpha2)*4... why is it so?
Please take the pain of reading this entire thing ...I would be really grateful if you help me out...thank you