A uniform clay disk with mass 3.5 kg and radius 0.52 m spins about its center of

Question

A uniform clay disk with mass 3.5 kg and radius 0.52 m spins about its center of mass with au angular velocity of 19 rad/a counter-clockwise. If the disk is reshaped (without friction) into a uniform thin hoop with a new radius of 0.90 m that rotates around its center of mass, then what is its new angular velocity? On a hydraulic lift, the small input piston is a circle with radius 0.050 m and the large output piston is a circle with radius of 2.5 rn. What force must be applied at the small input piston to balance a force of 2.0 x 10^3 N applied to the large output piston? A physical pendulum consists of a uniform thick hoop (of inner radius R^1 = 4.4 cm and outer radius R_o = 5.0 cm) hanging freely on a thin nail at its inner radius. The hoop is displaced by a small angle and released. What is the period of the resulting simple harmonic motion? A 0.32 kg mass is attached to a spring with a force constant of 46 N/m and is released a distance of 8.2 cm from the equilibrium position of the spring with an initial speed of 0.5 m/s toward its equilibrium position. What is the speed of the mass when it is at a distance of 1.6 cm. from the equilibrium position?

Explanation / Answer

Angular momentum = Moment of Inertia * Angular Velocity
Lin = I * w

Where I is moment of inertia of disk, = 1/2 * M*r^2
Lin = 1/2 * M*r^2 * w
Lin = 1/2 * 3.5 * 0.52^2 * 19
Lin = 9.0 kg*m2/s

Now,
Disk is reshaped into hoop
I = M*r^2
Lfi = 3.5 * 0.9^2 * w

Using Momentum Conservation -
Lin = Lfi
9.0 = 3.5 * 0.9^2 * w
w = 3.17 rad/s
New Angular Velocity, w = 3.17 rad/s

Please post seperate questions in seperate post.

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