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Snezhnost [94]
3 years ago
13

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.

Physics
2 answers:
hram777 [196]3 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]3 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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If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

  • Length of the massless beam;L = 4.00m
  • Distance of support from the left end; x = 3.00m
  • First mass; m1 = 31.3 kg
  • Distance of beam from  the left end( m₁ is attached to ); x_1 = ?
  • Second mass; m_2 = 61.7 kg
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Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

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u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

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a = acceleration

q = charge of proton

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E = 2.77*10^-18 / 1.6*10^-19

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