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

Respuesta :

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

Explanation: Please see the attachments below

Ver imagen Abdulazeez10
Ver imagen Abdulazeez10
Ver imagen Abdulazeez10

Answer:

930.21J

Explanation:

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

[tex]E_k=\frac{1}{2}I\omega^2[/tex]

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

The moment of inertia for a spherical shell is:

[tex]I=\frac{2}{3}MR^2[/tex]

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

[tex]I=\frac{2}{3}(8.05kg)(0.215m)^2=0.55\ kgm^2[/tex]

To calculate w we have to use the equation

[tex]\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}[/tex]

where we have taken w0=0 rad/s.

Finally, by replacing I and w we obtain:

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

HOPE THIS HELPS!