A block of mass m is connected to a string of negligible mass and slides on...

A block of mass m is connected to a string of negligible mass and slides on a slick surface in a circular path of radius 0.5 meters. The other end of the string passes through a hole in the surface and is connected to an identical block of mass m beneath the surface. All friction may be neglected in this problem and the string's mass and stretch are negligible.

1. Write the Newton's second law expressions for each block and the string with the apporpriate sum of physical forces and expressions for Fnet.

2 The top block is removed so that one block of mass m is in circular mtion with mass 3m pulling on the other end of the string. As the mass of 3m falls, pulling the string, the mass in circular motion speeds up in such a way that the prodcut of the radius and tangential speed remains constant RVtan = R(knot)V(knot)tan from [3]. Find the radius and spped at which the hanging mass reaches equilibrium.

Expert Solution

(1) For the mass which hangs
since that mass is not doing any motion in vertical direction therefore its acceleration will be zero.
So when we write equation it will be
T - mg = 0 where T is the tension in the string , m is the mass of the block and g is the acceleration due to gravity.
T = mg
Now foe the block which rotate in circualr path
Since it is in circualr motion therefore
T = mV²/R where T is the tension in the string and V is the velocity by which block is moving in circualr path and R is radius of circular path.
From first equation we know that T = mg
therfore on putting in the second we get
mg = mV²/R
V² = (gR)
(2) Now when hanging mass m is changed to 3m , when 3m mass get in equilibrium
Tension in the string will be
T = 3mg
And the mass m which is rotaing will have
T = mv²/r (Now the block is rotating with some other speed v and with some other radius r)
so
3mg = mv²/r
v² = 3gr
Since it is given that the velocity and product ratio remain constant therefore
VR = Constant
VR = vr
(gR)^1/2*R = (3gr)^1/2*r
Squaring both side we get
R³ = 3r³
r = R/(3)^1/3
SO intial radius R = 0.5 m is given therefore
r = 0.346 m
and new speed v = (3gr)^1/2 = 3.194 m/s

genius_generous answered 1 month ago

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