a disk with a radius of 0.1 m is spinning about its center with a constant angular speed of 10 rad/sec. What are the speed and magnitude of the acceleration of a bug clinging to the rim of the disk?

Respuesta :

The speed of the bug at the rim of the circular disk is [tex]\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}[/tex] and the acceleration of the bug is [tex]\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}[/tex].

Further Explanation:

Given:

The angular velocity of the disk is [tex]10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex].

The radius of the circular disk is [tex]0.1\,{\text{m}}[/tex].

Concept:

The linear speed of the bug present at the rim of the circular disk is given by.

[tex]v = r \times \omega[/tex]  

Here, [tex]v[/tex] is the linear speed,[tex]r[/tex] is the radius of the disk and [tex]\omega[/tex] is the angular speed of the disk.

Substitute [tex]0.1\,{\text{m}}[/tex] for [tex]r[/tex] and [tex]10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] for [tex]\omega[/tex] in above expression.

[tex]\begin{aligned}v &= 0.1\,{\text{m}} \times 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\\end{aligned}[/tex]  

The magnitude of the centripetal acceleration of the bug cling to the rim of the disk is given as.

[tex]{a_c} = \dfrac{{{v^2}}}{r}[/tex]  

Substitute [tex]1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] for [tex]v[/tex] and [tex]0.1\,{\text{m}}[/tex] for [tex]r[/tex] in above equation.

[tex]\begin{aligned}a_c&=\dfrac{(1\text{m/s})^2}{0.1\text{m}}\\&=\frac{1}{0.1}\text{m/s}^2\\&=10\,\text{m/s}^2\end{aligned}[/tex]  

Thus, the speed of the bug at the rim of the circular disk is [tex]\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}[/tex] and the acceleration of the bug is [tex]\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}[/tex].

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Answer Details:

Grade: College

Chapter: Uniform Circular motion

Subject: Physics

Keywords:  Circular disk, circular motion, angular speed, linear speed, bug clinging, rim of the disk, acceleration, magnitude, constant, rad/s.