A rigid body is spinning around an axis without any external torque. Due to change in temperature, the volume increases by $1\% $ . Its angular speed
A. Will increase approximately by $1\% $
B. Will decrease approximately by $1\% $
C. Will decrease approximately by $0.67\% $
D. Will increase approximately by $0.33\% $
Answer
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Hint: Here we have to first find the volume of the spherical body and then use the law of conservation of angular momentum to find what happens to the angular speed.
Complete step by step answer:The volume of a spherical body is given by
$V = \dfrac{4}
{3}\pi {R^3}$
Now, the change in radius and volume will be
$ \dfrac{{\Delta V}}
{V} = 3\frac{{\Delta R}}
{R} \\
\dfrac{{\Delta R}}
{R} = 0.33\% \\
$
The law of angular momentum conservation states that there is no change is angular momentum if there is no external torque acting on it.
It is given in the question that there is no external torque, so the body follows the law of angular momentum
So,
$L = I\omega $
Where $L$ is the angular momentum,
$I$ is the moment of inertia and $\omega $ is the angular velocity.
$ L = \dfrac{2}
{5}M{R^2}\omega \\
\dfrac{{\Delta L}}
{L} = 2\dfrac{{\Delta R}}
{R} + \dfrac{{\Delta \omega }}
{\omega } \\
$
As the magnitude remains constant, so $\dfrac{{\Delta L}}
{L} = 0$
Hence,
$
\dfrac{{\Delta \omega }}
{\omega } = - 2\dfrac{{\Delta R}}
{R} \\
= - 2 \times 0.33\% \\
= - 0.66\% \\
$
Hence, option C is correct.
The angular velocity decreases by $0.66\% $ .
Additional information:
If a system’s net external torque is zero, so the system’s angular momentum cannot shift. If a system’s net external torque is zero, then the system’s angular momentum remains the same.
The angular momentum changes according to the axis. We can conserve angular momentum by talking about an axis such that all forces travel around the axis or are parallel to it. Linear momentum however is independent of axis selection and cannot be maintained if the system has no-zero net power.
In addition to mass, shape and size of the body, the moment of inertia depends on the distribution of mass in the body along the axis of rotation.
Note:Here we have to pay attention to the negative sign. Since, the value is negative so the angular velocity will decrease otherwise it would increase.
Complete step by step answer:The volume of a spherical body is given by
$V = \dfrac{4}
{3}\pi {R^3}$
Now, the change in radius and volume will be
$ \dfrac{{\Delta V}}
{V} = 3\frac{{\Delta R}}
{R} \\
\dfrac{{\Delta R}}
{R} = 0.33\% \\
$
The law of angular momentum conservation states that there is no change is angular momentum if there is no external torque acting on it.
It is given in the question that there is no external torque, so the body follows the law of angular momentum
So,
$L = I\omega $
Where $L$ is the angular momentum,
$I$ is the moment of inertia and $\omega $ is the angular velocity.
$ L = \dfrac{2}
{5}M{R^2}\omega \\
\dfrac{{\Delta L}}
{L} = 2\dfrac{{\Delta R}}
{R} + \dfrac{{\Delta \omega }}
{\omega } \\
$
As the magnitude remains constant, so $\dfrac{{\Delta L}}
{L} = 0$
Hence,
$
\dfrac{{\Delta \omega }}
{\omega } = - 2\dfrac{{\Delta R}}
{R} \\
= - 2 \times 0.33\% \\
= - 0.66\% \\
$
Hence, option C is correct.
The angular velocity decreases by $0.66\% $ .
Additional information:
If a system’s net external torque is zero, so the system’s angular momentum cannot shift. If a system’s net external torque is zero, then the system’s angular momentum remains the same.
The angular momentum changes according to the axis. We can conserve angular momentum by talking about an axis such that all forces travel around the axis or are parallel to it. Linear momentum however is independent of axis selection and cannot be maintained if the system has no-zero net power.
In addition to mass, shape and size of the body, the moment of inertia depends on the distribution of mass in the body along the axis of rotation.
Note:Here we have to pay attention to the negative sign. Since, the value is negative so the angular velocity will decrease otherwise it would increase.
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