Recent content by Satvik Pandey

Oh! The numbers didn't worked as planned! I tried to find the the maximum value of velocity with which reduced mass can be projected so that make it's way up to the distance R from the origin. If it would projected with velocity(tangential) such that force acting on the reduced mass is...

Yes, the sphere is free to move. But we can covert this two body system into a single body system using reduced mass concept in order to solve these kinds of problem easily. :smile:

##R## is the radius of the sphere. If we consider radially inward direction to be positive then: with passage of time the separation reduced mass and origin decreases. Hence, there should be -ve sign with ##dx/dt## but ## \frac { d }{ dt } \left( V_{r} \right) ## should be positive because...

Yes, there ##a_{r}## stands for radial acceleration. . There should be -ve sign with ##dx/dt## because ##x## is decreasing with time but acceleration (radial) is ##dv_{r}/dt## and velocity(radial) is increasing with the passage of time there shouldn't be -ve sign. Am I right? Typo. Sorry :oops:

Homework Statement A particle is been acted by a central force exerted by a sphere of mass 'M' at a distance 'xo' initially from the particle (of same mass). At t=0 the particle has velocity Vt perpendicular to the line joining the particle and the center of the sphere. Find the time at which...

I too have same doubt. This question came in test papers of some coaching institution. And in the solution they simply wrote ##a=v^(2) /l.## I asked same question about the radial acceleration of ##M_{a}## here:(https://brilliant.org/discussions/thread/please-help-18/?ref_id=771297) but I...

So, ##I_{system}=\frac{ML^2}{12}+2 \left( m{ R }^{ 2 }+m{ \left( \frac { l }{ 2 } \right) }^{ 2 } \right) ## ##I_{w}=mR^{2}## So on putting these I got ##\omega_{0}=0.3v## Now time period of projectile is ##\frac { 2v\cos { \phi } }{ g } ## Time requires by wheels to rotate 538 deg...

Homework Statement We model a dirt bike as a rod of mass (M) 100 kg, and length (L) 2 m to which are attached two wheels of mass (m)=40 kg kg, and radius R=0.5 m. The bike goes off a spine ramp with angle ##\phi=1deg.## to the vertical on either side (see diagram below). Our question is, at...

Thank you!:-p

I revised the chapter of Center of Mass last night and I found that this can be solved even in a more easy way. Let us consider two blocks and the spring as a system. Net external force acting on the system is ##2F## So by Newton's second law acceleration of the CoM of the system ##...

Oh! I think I got it. Let us consider block 1,block 2 and spring as a system. So work done by all the forces is equal to the change in the kinetic and potential energy of the system. ##Fx+3Fy=\frac{k(x+y)^2}{2}+\frac{3mv^2}{2}## (extension would be maximum if velocities of blocks are...

I was making a mistake. I got that. The extension would be maximum at the moment when the relative velocity of the blocks along the line joining them would be zero.

If consider the other block to be moving rightward then Work done by all the forces on that block would be ##+3Fy-\frac{k(x+y)^2}{2}##.

Work done by all forces is equal to the change in the kinetic energy of the system. Taking block of mass ##2m## as a system. So Work done by F is ##+Fx## Work done by spring ##-\frac{k(x+y)^{2}}{2}## -ve sign because force exerted by the spring is in opposite direction of displacement. Change...

I separately applied work energy theorem for block1 and 2. At maximum compression velocities of the blocks along the line joining them should be equal but this is a case of maximum extension. I think the extension would be maximum when velocity of mass ##2m## is zero.