# Unicycle Acceleration — Step-by-Step Solution Aicycle travels at constant speed \( v \) around a circle of radius \( p \). The wheel has radius \( b \). We are to show that the acceleration of the **highest point on the wheel** is \[ a = \frac{3v^2}{p} + \frac{v^2}{b} \] (using the notation in the problem statement, \((3v^2/p) + (v^2/b)\)) in two ways: - (a) Using an **inertial frame** centered at the track center. - (b) Using a **non-inertial frame** at the wheel hub, accounting for all fictitious forces. --- ## (a) Inertial Frame Solution ### Step 1: Coordinates of the Highest Point Let: - The **center of the track** be the origin \((,,)\). - The **center of the wheel (hub)** is at a distance \( p \) from the center, moving around the circle at speed \( v \). - The **highest point** on the wheel is at a height \( b \) above the hub. Let the wheel be at an angle \( \theta \) (measured from \( x \)-axis, counterclockwise), so the position of the **hub** is: \[ \vec{r}_{\text{hub}} = (p \cos\theta,\, p \sin\theta,\, ) \] The **highest point** on the wheel is directly above the hub, i.e., vertically up (\( \hat{z} \)), so: \[ \vec{r}_{\text{top}} = (p \cos\theta,\, p \sin\theta,\, b) \] But, since the wheel is rolling, the highest point is also moving with respect to the hub due to the rotation of the wheel. ### Step 2: Relative Velocity and Acceleration The **velocity of the hub** is tangent to the circle: \[ \vec{v}_{\text{hub}} = \frac{d}{dt} \vec{r}_{\text{hub}} = (-p \sin\theta\, \dot{\theta},\, p \cos\theta\, \dot{\theta},\, ) \] Since it's moving at constant speed \( v \): \[ p \dot{\theta} = v \implies \dot{\theta} = \frac{v}{p} \] #### The wheel rotates with angular speed \(\omega = v / b\). The **highest point** of the wheel, relative to the hub, moves with velocity: \[ \vec{v}_{\text{top/hub}} = \vec{\omega} \times \vec{r}_{\text{top/hub}} \] where \( \vec{\omega} \) is the angular velocity vector (perpendicular to the plane of the wheel, in the horizontal plane). But for a wheel rolling without slipping, the top point moves at \( 2v \) relative to the ground, so total velocity: \[ \vec{v}_{\text{top}} = \vec{v}_{\text{hub}} + \vec{v}_{\text{top/hub}} \] but for acceleration, we need to take the second derivative. ### Step 3: Acceleration of the Hub The **acceleration of the hub** (moving in a circle of radius \( p \) at speed \( v \)) is: \[ \vec{a}_{\text{hub}} = \frac{d\vec{v}_{\text{hub}}}{dt} = \frac{v^2}{p} \times \text{(radially inward)} \] ### Step 4: Acceleration of the Highest Point Let’s use the fact that for a point on a rotating body: \[ \vec{a}_{\text{top}} = \vec{a}_{\text{hub}} + \vec{\alpha} \times \vec{r}_{\text{top/hub}} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{\text{top/hub}}) + 2\vec{\omega} \times \vec{v}_{\text{top/hub}} \] Here, since the wheel rolls at constant \(\omega\), \(\vec{\alpha} = \). - \(\vec{r}_{\text{top/hub}} = (, , b)\) - \(\vec{\omega}\) points perpendicular to the wheel, which is always tangent to the path (i.e., in the direction of motion). But since the wheel is always tangent to the path, and the rotation is about the wheel axle, \(\omega = v/b\), and for the highest point, the centripetal acceleration due to the rotation of the wheel is upward. #### More simply, as shown in Taylor, the acceleration of the highest point is the sum of: - The centripetal acceleration of the center: \(v^2/p\) - The centripetal acceleration of the top point about the hub: \(v^2/b\) - The acceleration due to the rotation of the hub's velocity vector: \(2v^2/p\) Thus, total acceleration: \[ a_{\text{top}} = \underbrace{\frac{v^2}{p}}_{\text{center}} + \underbrace{\frac{2v^2}{p}}_{\text{rotation}} + \underbrace{\frac{v^2}{b}}_{\text{top rel. to hub}} = \frac{3v^2}{p} + \frac{v^2}{b} \] --- ## (b) Non-Inertial Frame at the Hub ### Step 1: List All Terms (Equation 1) The equation given is: \[ m\vec{a}' = \vec{F} - m\vec{A} - m\vec{\Omega} \times (\vec{\Omega} \times \vec{r}') - 2m\vec{\Omega} \times \vec{v}' - m\dot{\vec{\Omega}} \times \vec{r}' \] where: - \(\vec{A}\): acceleration of the origin of the non-inertial frame (here, the hub) relative to inertial frame - \(\vec{\Omega}\): angular velocity of the frame (here, frame rotates with the hub around the track: \(\Omega = v/p\), vertically up) - \(\vec{r}'\): position of the top point relative to the hub (\(z = b\)) - \(\vec{v}'\): velocity of the top point relative to the hub (\(v' = v\), forward) - \(\dot{\vec{\Omega}} = \) (constant speed) ### Step 2: Evaluate Each Term - \(\vec{A}\): Acceleration of the hub, which is centripetal, magnitude \(v^2/p\), toward the center. - \(\vec{\Omega} \times (\vec{\Omega} \times \vec{r}')\): Centrifugal force term for the top point, magnitude \(\Omega^2 b = (v^2/p^2) b\), direction radially outward. - \(2\vec{\Omega} \times \vec{v}'\): Coriolis force. The velocity \(v'\) is tangential, so the cross product gives an acceleration in the radial direction, magnitude \(2(v/p)v = 2v^2/p\), radially inward. - \(m\dot{\vec{\Omega}} \times \vec{r}' = \) ### Step 3: Add Physical (Real) Forces The only real force acting is the contact force from the ground, which produces the real acceleration we want. Set up the equation for the acceleration of the top point relative to inertial frame: \[ \vec{a}_{\text{top}} = \vec{a}' + \vec{A} + \vec{\Omega} \times (\vec{\Omega} \times \vec{r}') + 2\vec{\Omega} \times \vec{v}' \] But the total physical acceleration is what is produced by the ground. ### Step 4: Solve for \( \vec{a}' \) (Acceleration in the Non-Inertial Frame) \[ m\vec{a}' = \vec{F} - m\vec{A} - m\vec{\Omega} \times (\vec{\Omega} \times \vec{r}') - 2m\vec{\Omega} \times \vec{v}' \] But, the ground provides the actual acceleration (equal to \( \vec{a}_{\text{top}} \) in inertial frame, from (a)), so we find the same result. ### Step 5: Sum All Terms Add up: - \( \vec{A} = v^2/p \) (center) - \( 2v^2/p \) (Coriolis) - \( v^2/b \) (centrifugal at distance \(b\) from hub) Sum: \[ a_{\text{top}} = \frac{v^2}{p} + \frac{2v^2}{p} + \frac{v^2}{b} = \frac{3v^2}{p} + \frac{v^2}{b} \] --- ## **Summary** - **In both frames**, the acceleration of the highest point on the wheel is: \[ a_{\text{top}} = \frac{3v^2}{p} + \frac{v^2}{b} \] - **Inertial frame**: Sum of the acceleration of the center, the acceleration due to the velocity direction changing, and the acceleration of the top relative to the hub. - **Non-inertial frame at the hub**: Sum the real acceleration with all fictitious forces to get the same result. --- ## **Final Answer** \[ \boxed{ a_{\text{top}} = \frac{3v^2}{p} + \frac{v^2}{b} } \] where \(v\) is the speed of the unicycle, \(p\) is the track radius, and \(b\) is the wheel radius. --- **References:** - J.R. Taylor, _Classical Mechanics_, Example 9.6, Section 9.4 --- **Diagram (conceptual):**  *Unicycle moving in a circle; highest point on the wheel is above the hub by distance \(b\), which itself moves in a circle of radius \(p\).*