Riders in a carnival ride stand with their backs against the wall of a circular room of diameter 8.0 m. The room is spinning horizontally about an axis through its center at a rate of 45 rev/min when the floor drops so that it no longer provides any support for the riders. What is the minimum coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall? • Nerd Notes

First, convert the rotational speed to radians per second: $\omega = 45 , \frac{\text{rev}}{\text{min}} \times \frac{1 , \text{min}}{60 , \text{s}} \times \frac{2\pi , \text{radians}}{1 , \text{rev}}$ .

Note you can convert to linear velocity $v$ instead of using $\omega$ , since $v = \omega r$ , where $r$ is the radius of the circle.

The minimum coefficient of static friction is found using the balance of forces in the vertical direction, with the static frictional force providing the upward force to counteract gravity, and the centripetal force being provided by the normal force.

Step	Formula / Derivation	Reasoning
1	$a_c = \frac{v^2}{R}$	Centripetal acceleration formula.
2	$v = R\omega$	Linear velocity in terms of radius and angular velocity.
3	$a_c = R\omega^2$	Substitute $v$ into the centripetal acceleration formula.
4	$F_c = m a_c = m R\omega^2$	Centripetal force provided by the normal force.
5	$f_s = \mu_s N$	Static frictional force opposing gravity.
6	$f_s = m g$	Static frictional force equals gravitational force to prevent sliding.
7	$\mu_s = \frac{f_s}{N} = \frac{mg}{mR\omega^2}$	Substitute $f_s$ and $N$ with their equivalents.
8	$\mu_s = \frac{g}{R\omega^2}$	Cancel $m$ from the equation.
9	$\mu_s = \frac{g}{R\left(\frac{45}{60} \times 2\pi\right)^2}$	Substitute $\omega$ with its value in terms of rev/min to rad/s.
10	$\mu_s = \frac{9.8}{4\left(\frac{3}{4} \pi\right)^2}$	Substitute given values for $g$ and $R$ .
11	$\mu_s = \frac{9.8}{4\left(\frac{9}{16} \pi^2\right)}$	Simplify the expression for $\omega^2$ .
12	$\boxed{\mu_s = \frac{9.8 \times 16}{4 \times 9 \pi^2}}$	Final expression for $\mu_s$ .

Now let’s calculate the exact value for the minimum coefficient of static friction required.

Kinematics	Forces
$\Delta x = v_i t + \frac{1}{2} at^2$	$F = ma$
$v = v_i + at$	$F_g = \frac{G m_1m_2}{r^2}$
$a = \frac{\Delta v}{\Delta t}$	$f = \mu N$
$R = \frac{v_i^2 \sin(2\theta)}{g}$

Circular Motion	Energy
$F_c = \frac{mv^2}{r}$	$KE = \frac{1}{2} mv^2$
$a_c = \frac{v^2}{r}$	$PE = mgh$
	$KE_i + PE_i = KE_f + PE_f$

Momentum	Torque and Rotations
$p = m v$	$\tau = r \cdot F \cdot \sin(\theta)$
$J = \Delta p$	$I = \sum mr^2$
$p_i = p_f$	$L = I \cdot \omega$

Constant	Description
$g$	Acceleration due to gravity, typically $9.8 , \text{m/s}^2$ on Earth’s surface
$G$	Universal Gravitational Constant, $6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2$
$\mu_k$ and $\mu_s$	Coefficients of kinetic ( $\mu_k$ ) and static ( $\mu_s$ ) friction, dimensionless. Static friction ( $\mu_s$ ) is usually greater than kinetic friction ( $\mu_k$ ) as it resists the start of motion.
$k$	Spring constant, in $\text{N/m}$
$M_E = 5.972 \times 10^{24} , \text{kg}$	Mass of the Earth
$M_M = 7.348 \times 10^{22} , \text{kg}$	Mass of the Moon
$M_M = 1.989 \times 10^{30} , \text{kg}$	Mass of the Sun

Variable	SI Unit
$s$ (Displacement)	$\text{meters (m)}$
$v$ (Velocity)	$\text{meters per second (m/s)}$
$a$ (Acceleration)	$\text{meters per second squared (m/s}^2\text{)}$
$t$ (Time)	$\text{seconds (s)}$
$m$ (Mass)	$\text{kilograms (kg)}$

Simple Harmonic Motion
$F = -k x$
$T = 2\pi \sqrt{\frac{l}{g}}$
$T = 2\pi \sqrt{\frac{m}{k}}$

Variable	Derived SI Unit
$F$ (Force)	$\text{newtons (N)}$
$E$ , $PE$ , $KE$ (Energy, Potential Energy, Kinetic Energy)	$\text{joules (J)}$
$P$ (Power)	$\text{watts (W)}$
$p$ (Momentum)	$\text{kilogram meters per second (kgm/s)}$
$\omega$ (Angular Velocity)	$\text{radians per second (rad/s)}$
$\tau$ (Torque)	$\text{newton meters (Nm)}$
$I$ (Moment of Inertia)	$\text{kilogram meter squared (kgm}^2\text{)}$
$f$ (Frequency)	$\text{hertz (Hz)}$

Prefix	Symbol	Power of Ten
Pico-	p	$10^{-12}$
Nano-	n	$10^{-9}$
Micro-	µ	$10^{-6}$
Milli-	m	$10^{-3}$
Centi-	c	$10^{-2}$
Deci-	d	$10^{-1}$
(Base unit)	–	$10^{0}$
Deca- or Deka-	da	$10^{1}$
Hecto-	h	$10^{2}$
Kilo-	k	$10^{3}$
Mega-	M	$10^{6}$
Giga-	G	$10^{9}$
Tera-	T	$10^{12}$

Everything you need to master Physics — for free.

Meet Phy. The world's most advanced personal AI tutor.

Learning supercharged. 1000+ Physics problems + AI help.

7 AP Physics 1 units condensed into 8 short articles.

Proven tip, tricks, and hacks to help you score that coveted five.

Learn Physics, from scratch at your own pace.

Guaranteed performance boost in scores and understanding.

5 weeks for a 5. Expedited progress. Stellar Results.

1-to-1 lessons covering AP Physics from start to end.

Other 1-to-1 and group programs to accelerate Physics mastery.

AP Physics

Unit 3 - Circular Motion

Advanced

Mathematical

GQ

You're a Phy Pro Member

Supercharge UBQ with

Need Help? Ask Phy To Explain This Problem

Topics in this question

See how Others Did on this question | Coming Soon

Related Questions

5 More similar questions to the one you just solved

Discussion Threads

Leave a Reply Cancel reply

Nerd Notes

Discover the world's best Physics resources

Continue with

Sign In to View Your Questions

Share This Question

Made By Nerd-Notes.com

Phy Pro

$11.99

Error Report

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

You can close this ad in 7 seconds.

Ads display every few minutes. Upgrade to Phy Pro to remove ads.

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

How to UBQ

1. UBQ Credits

2. UBQ Stats

3. Point System

Your learning experience, set by your pace

4. Take your time!

Phy gives feedback.

4. Feedback

Jason here! Feeling uneasy about your next physics test? We will help boost your grade in just two hours.

We use site cookies to improve your experience. By continuing to browse on this website, you accept the use of cookies as outlined in our privacy policy.