Question

A point on a wheel of radius 40 cm that is rotating at a constant 5.0 revolutions per second is located 0.20 m from the axis of rotation. What is the acceleration of that point due to the spin of the wheel?

Answer 1

Answer:

$197.2 m/s^2$

Explanation:

The centripetal acceleration of a point moving by circular motion is given by:

$a=\omega^2 r$

where

$\omega$ is the angular velocity

r is the distance from the axis of rotation

The point on the wheel makes 5.0 revolutions per second, so the frequency is

$f=\frac{5}{1}=5 Hz$

and the angular velocity is

$\omega=2\pi f = 2\pi (5)=31.4 rad/s$

While the distance of the point from the axis of rotation is

$r=0.20 m$

Substituting, we find the acceleration:

$a=(31.4)^2(0.20)=197.2 m/s^2$

Answer 2

The acceleration of that point due to the spin of the wheel is about 2.0 × 10² m/s²

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Further explanation

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

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Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

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Given:

radius of wheel = R = 40 cm = 0.4 m

angular velocity = ω = 5.0 rev/s = 10π rad/s

distance of the point from axis of rotation = r = 0.20 m

Asked:

acceleration of that point = a = ?

Solution:

$a = \omega^2 r$

$a = (10 \pi )^2 \times 0.20$

$a = 20 \pi^2 \texttt{ m/s}^2$

$a \approx 2.0 \times 10^2 \texttt{ m/s}^2$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion