Question

A 1.2-m-long massless rod is pivoted at one end and swings around in a circle on a frictionless table. A block with a hole through the center can slide in and out along the rod. Initially, a small piece of wax holds the block 20 cm from the pivot. The block is spun at 40 rpm , then the temperature of the rod is slowly increased. When the wax melts, the block slides out to the end of the rod. What is the final angular velocity?

Answer #1

The moment of inertia (I) for the block (which may be considered as
a point particle, in fact it must be considered as one because we
are not given its dimensions) is mr². There are two values for I,
one where the block is a distance of 20cm (0.20m) from the pivot,
and one that is 1.2m from the pivot. Therefore:

------------------>mass divides out

= (0.20m)²(40rev/mim) / (1.2m)²

= 1.11 rev/min

If this should be in SI units, then:

1.1rpm = 1.1rev/min(2πrad/1.0rev)(1.0min/60s)

= 0.115rad/s

A 1.2-m-long massless rod is pivoted at one end and swings
around in a circle on a frictionless table. A block with a hole
through the center can slide in and out along the rod. Initially, a
small piece of wax holds the block 40 cm from the pivot. The block
is spun at 50 rpm , then the temperature of the rod is slowly
increased. When the wax melts, the block slides out to the end of
the rod....

A uniform rod of mass M and length L is pivoted at one
end. The rod is left to freely rotate under the influence of its
own weight. Find its angular acceleration α when it makes
an angle 30° with the vertical axis. Solve for M=1 Kg, L=1 m,
take g=10 m s-2. Hint: Find the center of mass for the rod, and
calculate the torque, then apply Newton as τ= Ι·α

A uniform rod of mass M and length L is pivoted at one end. The
rod is left to freely rotate under the influence of its own weight.
Find its angular acceleration α when it makes an angle 30° with the
vertical axis. Solve for M=1 Kg, L=1 m, take g=10 m s-2. Your
answer in X.X rad s-2. Hint: Find the center of mass for the rod,
and calculate the torque, then apply Newton as τ= Ι·α

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