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2 years ago

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A hollow spherical shell has mass 8.05kg and radius 0.215m . It is initially at rest and then rotates about a stationary axis th

at lies along a diameter with a constant acceleration of 0.895rad/s2 .
What is the kinetic energy of the shell after it has turned through 5.25rev ?

Express your answer with the appropriate units.

2 answers:

hram777 [196]2 years ago

4 0

Answer: The kinetic energy of the shell after it has turned through 5.25rev is 7.32J

Explanation: Please see the attachments below

Neporo4naja [7]2 years ago

3 0

Answer:

930.21J

Explanation:

The kinetic energy of a solid body that is rotating, is given by:

$E_k=\frac{1}{2}I\omega^2$

where I is the moment of inertia and w is the angular velocity.

The moment of inertia for a spherical shell is:

$I=\frac{2}{3}MR^2$

where M is the mass and R is the radius of the sphere. By replacing we have

$I=\frac{2}{3}(8.05kg)(0.215m)^2=0.55\ kgm^2$

To calculate w we have to use the equation

$\omega^2=\omega_0^2+2\alpha \theta\\\\\omega=\sqrt{2(0.895rad/s^2)(5.25rev*\frac{360\°}{1\ rev})}=58.16\frac{rad}{s}$

where we have taken w0=0 rad/s.

Finally, by replacing I and w we obtain:

$E_k=\frac{1}{2}I\omega^2 =\frac{1}{2}(0.55\ kgm^2)(58.16\frac{rad}{s})^2=930.21J$

HOPE THIS HELPS!

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