Constraint Rotation

Classical Mechanics Level 2

A uniform rod of mass m and length 2l lies on a smooth horizontal surface. A particle of mass m is connected to a string of length l whose other end is connected to the rod. Initially the string is taut and both rod and string lies in the same horizontal plane with $90^\circ$ angle between them If the particle is given initially a velocity v perpendicular to the string then just after giving velocity v to the particle...

Angular velocity of the rod will be

\frac{v}{2l}

Tension in the string will be

\frac{mv^{2}}{5l}

Angular acceleration of the rod will be

\frac{6v^{2}}{5l}

linear acceleration of the centre of mass of the rod will be

\frac{v^{2}}{4l}

2 solutions

Jatin Yadav
Mar 19, 2014

On rod:

$T = ma$

$Tl = \frac{m l^2}{3} \alpha$

On particle,

$T = m(\frac{v^2}{l} - l \alpha - a)$

Solve to get:

$a = \frac{v^2}{5l}$

$\alpha = \frac{3v^2}{5l^2}$

$T = \frac{mv^2}{5l}$

I can't see how you come to the equation for the particle. Can you help me out, please?

Tom Van Lier - 7 years, 2 months ago

I have been looking at the problem for some days now and my reasoning can still be wrong, but aren't you adding vectors which are perpendicular (normal end tangential) without applying Pythagoras?

Tom Van Lier - 7 years, 2 months ago

it is because he is writing equation of motion for particle with respect to a point on the rod where string is connected. rod is in combined motion ( both rotation and translation ) with Acm , and angular acceleration of alpha . acceleration of that point is Acm + L(alpha) . now you can either use pseudo force concept or you can use relative acceleration . path of particle with respect to point is circular.

Nitin Sachan - 6 years, 9 months ago

Even mee to

Sandaparthi Venkata Trinadha Murty - 3 years, 3 months ago

On rod how tension =ma how did sought to write?

raja sharma - 3 years, 2 months ago

Sid Leo
Apr 11, 2014

no his equations are right......consider de frame of refrence to be de point where the rod nd thread are attached........dat point is aceelarating wid [l][alpha]+a in downward direction .......wid respect to dat refrence de bob or de particle undergoes circular motion.......soo consider pseudo force on de particle and write its centripetal acceleration equation....so yu get T+ml[alpha] +ma=mv^2/l......dts hw he gt it if i am nt wrong.....

Shouldn't the torque equation be ml^2/12 about com

Tanmay Bhoite - 7 years, 1 month ago

The length of rod is 2 l so I about COM is m (2l)^2/13=ml^2/3

Kapil Sethi - 7 years, 1 month ago

if the row has length 2l then moment of inertia about COM should be 4 m l^2/3?

Kapil Sethi - 7 years, 1 month ago

Shuould n't Tl = ml^2/12*(alpha)

Usama Khidir - 5 years, 9 months ago

Constraint Rotation

2 solutions

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