Posted by **shosho** on Thursday, November 5, 2009 at 5:51am.

A puck of mass m = 1.5 kg slides in a circle of radius r = 20 cm on a frictionless table while attached to a hanging cylinder of mass M = 2.5 kg by a cord through a hole in the table. What is the speed of mass m that keeps the cylinder at rest?

- phisics -
**bobpursley**, Thursday, November 5, 2009 at 11:13am
Mg =mv^2/r

Repost if you need your work critiqued.

- phisics -
**Kevin**, Tuesday, April 10, 2012 at 8:41am
To keep the cylinder in equilibrium both side must be equal.

(m1)*(v^2/r)=(m2)*g Left side=Right side

(v^2/r) = (m2)*g / m1 isolate “v” on the left…

v^2 = (m12)*g*r/m1

v = sqrt((m2)*g*r /(m1)) plug in the values and you are done…

v = sqrt(2.5*9.81*.2)/(1.5) = 1.80831

Dr. Cooper will be proud of you.

m1 = mass of the puck = 1.5 km

m2 = mass of the cylinder = 2.5 km

r= radius = 20 cm = .2 m (SI units)

v = velocity

g = 9.81 (if do not know what “g” is you should look it up)

sqrt = square root

- physics -
**Al**, Tuesday, September 25, 2012 at 7:55pm
v = sqrt((2.5*9.81*.2)/(1.5)) don't forget the parentheses. You will get seriously messed up haha

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