a uniform solid cylinder of mass m and radius r is released from the top of a rough inclined plane of inclination theta and length of slant = l . the coefficient of friction is muu=(1/4)tan theta .find time taken by cyillender to reach the bottom , total kinetic energy of the cylinder at the bottom , total work done by friction.
pls reply fast...
Answers
Answered by
1
suppose there is a frictional force of f=(mu)N.
where N=normal reaction. now
if there us a friction force there there will be a torque acting on the body.
let ,torque=t.
THEN,t=f.r where r =radius of the solid cylinder.
Then f=mg(cos€)×(mu), where €=theta.
So we know ,
t=I×@.where I and @ represents the moment of inertia and the angular acceleration of the cylinder.
as I=0.5×m×r².
now considering the linear motion, we have,
mg(sin€)-f=ma,where a=acceleration.
Also, f=Ia/r².
Then, ma=mgsin€-Ia/r²
=>a[1+I/0.5mr²]=gsin€
where g=acceleration due to gravity.
a=(gsin€)/(1+0.5)=gsin€/1.5 m/s².
now solving we get, a=(2/3)gsin€
now ,applying energy conservation law,
v=[√2gh/1+k²/r²].
where k= radius of gyration of the body.
where N=normal reaction. now
if there us a friction force there there will be a torque acting on the body.
let ,torque=t.
THEN,t=f.r where r =radius of the solid cylinder.
Then f=mg(cos€)×(mu), where €=theta.
So we know ,
t=I×@.where I and @ represents the moment of inertia and the angular acceleration of the cylinder.
as I=0.5×m×r².
now considering the linear motion, we have,
mg(sin€)-f=ma,where a=acceleration.
Also, f=Ia/r².
Then, ma=mgsin€-Ia/r²
=>a[1+I/0.5mr²]=gsin€
where g=acceleration due to gravity.
a=(gsin€)/(1+0.5)=gsin€/1.5 m/s².
now solving we get, a=(2/3)gsin€
now ,applying energy conservation law,
v=[√2gh/1+k²/r²].
where k= radius of gyration of the body.
Similar questions