Interactive physics simulator
Linear & Angular Velocity Relation
Explore how rotational speed and position dictate linear velocity. Analyze the fundamental relation $v = r\omega$, observe how tangential vectors grow with radius, and simulate static friction limits on a rotating turntable.
Linear & Angular Velocity Relation Lab
Interact with parameters on the right and click Simulate to play rotation frames or watch graphs.
Live Telemetry
- Angular Speed (ω)
- 0.00 rad/s
- Radius (r)
- 0.00 m
- Tangential Speed (v)
- 0.00 m/s
- Req Centrip Accel
- 0.00 m/s²
- Friction Threshold
- 0.00 m/s²
- Turntable Slip State
- Stable
Introduction to Linear and Angular Velocity Relation
When a wheel rotates, every single point on that wheel turns through the exact same angle in the same amount of time. In other words, every point shares the same **angular velocity** (ω). However, do they all travel at the same **linear speed**? If you watch a point near the axle compared to a point on the outer tread, the outer point sweeps out a much larger circumference in that same time interval. This means the linear tangential speed ($v$) of any point on a rotating object depends on its distance from the axis of rotation ($r$).
Core Derivation & Formulas
1. Derivation of v = r · ω
Recall the definition of an angle in radians: the angle θ (in radians) is defined as the arc length (s) divided by the radius (r) of the circular path:
If we take the derivative of both sides of this equation with respect to time (t) for a constant radius (r), we get:
Since **ds/dt** is the linear tangential speed (v) along the circular arc, and **dθ/dt** is the angular velocity (ω) of the body, we arrive at the fundamental relationship:
2. Tangential Velocity Vector Direction
Although speed (v) might remain constant in uniform circular motion, **velocity is a vector** and includes direction. At any instant, the linear velocity vector of a rotating particle points **tangent to its circular trajectory**. It is always perpendicular to the radial connector line. If the constraint holding the particle in circular motion (like a string, or friction) is suddenly cut, the particle flies off along this tangent line due to inertia.
3. Centripetal Acceleration & Friction Thresholds
To keep an object moving in a circle, it must accelerate toward the center. This is called **centripetal acceleration** (ac):
On a flat rotating turntable, this centripetal acceleration is supplied entirely by **static friction** between the object and the turntable surface. The maximum acceleration friction can supply is:
If the required centripetal acceleration exceeds this threshold (i.e. ω²r > μsg), static friction breaks away, and the object **slips**, sliding off the turntable in a tangent straight line.
Solved Numerical Examples
A playground carousel of radius 3.5 meters rotates at a constant rate of 12.0 RPM (revolutions per minute). Determine: (a) its angular velocity in radians per second, and (b) the tangential linear speed of a child sitting on the outer edge.
View Step-by-Step Solution
- Given: Radius r = 3.5 m, rotational speed = 12.0 RPM.
- Convert the rotational speed from RPM to radians per second (ω):
1 rev = 2π rad, 1 min = 60 s.
ω = 12.0 × (2π / 60) = 0.4π rad/s ≈ 1.257 rad/s. - Use the relationship between linear speed and angular speed: v = r · ω.
- Substitute the values: v = 3.5 m × 1.257 rad/s ≈ 4.40 meters per second.
- Results: The carousel rotates at 1.26 rad/s, carrying the child along the perimeter at a linear tangential speed of 4.40 m/s.
A helicopter rotor has blades that are 6.0 meters long and spin at 350 RPM. Find the linear tangential speed of: (a) the tip of the blade, and (b) a point along the blade located 2.0 meters from the center hub.
View Step-by-Step Solution
- Given: Total length (radius of tip) Rtip = 6.0 m, radius of inner point Rinner = 2.0 m, rotation rate = 350 RPM.
- Convert 350 RPM to angular velocity (ω):
ω = 350 × (2π / 60) = (35/3)π rad/s ≈ 36.65 rad/s. - Recall that all parts of a rigid rotating body share the exact same angular velocity (ω = 36.65 rad/s).
- Calculate linear speed at the tip: vtip = Rtip · ω.
vtip = 6.0 m × 36.65 rad/s ≈ 219.9 meters per second. - Calculate linear speed at the inner point: vinner = Rinner · ω.
vinner = 2.0 m × 36.65 rad/s ≈ 73.3 meters per second. - Results: The tip of the rotor moves at a high speed of 220 m/s, whereas the point closer to the hub moves at only 73.3 m/s.
A coin is placed on a horizontal rotating turntable at a distance of 15.0 cm from the center axle. If the coefficient of static friction between the coin and the surface is μ<sub>s</sub> = 0.40, calculate: (a) the maximum centripetal acceleration the coin can experience without slipping, and (b) the maximum angular speed in both rad/s and RPM at which the turntable can spin before the coin slides off. (Use g = 9.80 m/s²).
View Step-by-Step Solution
- Given: Radius r = 15.0 cm = 0.15 m, static friction coefficient μs = 0.40, gravitational acceleration g = 9.80 m/s².
- The static friction force provides the necessary centripetal force: fs = m·ac ≤ μs·m·g.
- Calculate the maximum centripetal acceleration (ac,max):
ac,max = μs · g = 0.40 × 9.80 m/s² = 3.92 m/s². - Relate centripetal acceleration to angular velocity: ac = ω²·r ⇒ ω²·r ≤ ac,max.
- Solve for maximum angular velocity (ωmax):
ωmax = √(ac,max / r) = √(3.92 / 0.15) = √(26.133) ≈ 5.11 rad/s. - Convert angular velocity to RPM:
RPM = 5.11 × (60 / 2π) ≈ 48.8 RPM. - Results: The maximum centripetal acceleration is 3.92 m/s², and the turntable can spin up to 5.11 rad/s (48.8 RPM) before slip occurs.
Conceptual Practice
If all parts of a spinning turntable share the same angular velocity, why do points further from the center have a larger linear velocity? Explain using the definition of linear distance.
Show Explanation
Linear speed ($v$) is distance traveled per unit time. During one full revolution ($T$), a point at radius $r$ travels a path length equal to the circumference $2pi r$. Although all points complete one rotation in the same period $T$ (constant $omega$), points at a larger radius must trace out a much larger circle ($2pi r$) in that same time. Thus, they must travel faster. This is expressed by $v = rac{2pi r}{T} = romega$, showing linear velocity increases proportionally with distance from the center.
A car drives around a circular track. If it doubles its linear speed, what happens to the required centripetal acceleration? How does this relate to the traction limits of tires?
Show Explanation
The required centripetal acceleration is given by $a_c = v^2/r$. If the linear speed $v$ is doubled while the radius $r$ remains constant, the centripetal acceleration increases by a factor of four ($2^2 = 4$). Since static friction between tires and road provides this centripetal acceleration ($f_s = m a_c$), doubling your speed requires four times the friction force to keep the car on the track. If this exceeds the friction limit ($mu_s g$), the car will skid.
Explain how a compact disc (CD) player uses the relation $v = romega$ during playback. Why does the motor slow down as the laser reading head moves from the inner tracks to the outer edge?
Show Explanation
A CD player is designed to read data at a constant linear speed (Constant Linear Velocity, or CLV) of about 1.2 to 1.4 m/s. According to $v = romega$, to keep $v$ constant as the reading head moves outward (larger radius $r$), the angular speed $omega$ must decrease. Therefore, the disc spins fastest (~500 RPM) when reading data near the inner tracks and slows down (~200 RPM) as it reads the outer tracks.
A rock tied to a string is whirled in a horizontal circle at a constant angular speed. If the string suddenly snaps, describe the path the rock takes. What physical law dictates this behavior?
Show Explanation
The moment the string snaps, the centripetal force holding the rock in a circle drops to zero. According to Newton's First Law of Motion (Inertia), an object in motion will continue in a straight line at a constant speed unless acted on by an external force. Since the rock's instantaneous velocity vector points tangent to the circle, the rock will fly off in a straight line **tangent to the circular path** at the snap point (ignoring the gradual downward pull of gravity).
Frequently Asked Questions
What is the relationship between linear speed and angular speed?
The linear tangential speed (v) of a point on a rotating body is directly proportional to its distance from the axis of rotation (r) and its angular speed (ω), expressed as v = r · ω.
What units must be used in the formula v = rω?
To get linear speed in meters per second (m/s), the radius (r) must be in meters (m) and the angular speed (ω) must be in radians per second (rad/s). Revolutions or degrees must be converted to radians.
Why do points at different radii on a rigid wheel have different linear speeds?
Because all points complete one full rotation in the same time, but points further from the center must travel around a larger circumference. To cover a larger distance in the same time, they must move at a faster linear speed.
What is the difference between tangential speed and angular speed?
Tangential speed (v) measures the linear distance covered per second by a point along the circular arc. Angular speed (ω) measures the angle swept out per second by the rotating body as a whole.
How do you convert RPM to radians per second?
Multiply the RPM value by 2π to convert revolutions to radians, and divide by 60 to convert minutes to seconds. Simplified: ω = RPM × (π / 30).
If a wheel is spinning, does the center axle have a linear speed?
No. At the exact center of rotation, the radius r is zero. Substituting r = 0 into v = rω yields a linear speed of zero, even though it shares the same angular speed.
What is centripetal acceleration, and how does it relate to linear speed?
Centripetal acceleration is the rate of change of direction of the velocity vector, pointing toward the center of rotation. It is related to linear speed by a<sub>c</sub> = v²/r, or to angular speed by a<sub>c</sub> = ω²r.
What causes an object on a rotating turntable to slip?
An object slips when the required centripetal acceleration (a<sub>c</sub> = ω²r) exceeds the maximum static acceleration that friction can provide (a<sub>f,max</sub> = μ<sub>s</sub>g).
Does mass affect the speed at which an object slips off a turntable?
No. Centripetal force is mω²r and friction force is μ<sub>s</sub>mg. Setting them equal gives mω²r = μ<sub>s</sub>mg, where mass (m) cancels out on both sides, leaving ω²r = μ<sub>s</sub>g.
In what direction does an object fly when it slips off a spinning wheel?
It flies off in a straight line tangent to the circular path at the exact position it breakaway, moving in the direction of its instantaneous linear velocity.
How do gear ratios utilize the relation v = rω?
When two gears mesh, their outer teeth touch and share the exact same tangential linear speed (v). Therefore, r<sub>1</sub>ω<sub>1</sub> = r<sub>2</sub>ω<sub>2</sub>, meaning the gear with a smaller radius must rotate with a higher angular speed.
Why do wind turbine tips spin so fast?
Because wind turbine blades are very long (large radius r). Even at a low angular speed of 10-15 RPM, the large radius results in a very high linear speed at the blade tips, often exceeding 70 m/s.