Rotational kinematics — tangential (linear) acceleration: A body is rotating along a circular path of radius r with an angular acceleration α (in rad/s^2). What is the expression for its tangential (linear) acceleration a along the path?

Introduction / Context:Tangential (linear) acceleration connects rotational motion with linear motion at a radius. In mechanical and civil engineering problems involving pulleys, gears, flywheels, or curved vehicle paths, converting angular acceleration to linear acceleration is a routine step.

Given Data / Assumptions:

• Circular motion of radius r (meters).
• Angular acceleration α (rad/s^2) is constant or instantaneous.
• Tangential acceleration a is the linear acceleration along the circle.

Concept / Approach:By kinematics, linear and angular measures are related through the radius. The linear velocity is v = ω * r and the tangential acceleration is the time rate of change of v due to changing angular speed, so a_t depends on α and r directly.

Step-by-Step Solution:Start from v = ω * r.Differentiate with respect to time: a_t = dv/dt = r * dω/dt.But dω/dt = α, hence a_t = r * α.Therefore, a = α * r.

Verification / Alternative check:Units: α has units rad/s^2 (radian is dimensionless), multiplied by r (m) gives m/s^2, which are correct units for linear acceleration.

Why Other Options Are Wrong:

• a = α / r: Inverts the dependence; dimensions become 1/(s^2·m), not m/s^2.
• a = r / α: Also dimensionally incorrect and physically implausible.
• none of these: Incorrect because a = α * r is the standard relation.

Common Pitfalls:

• Mixing tangential acceleration a_t = α * r with normal (centripetal) acceleration a_n = v^2 / r = ω^2 * r; both exist simultaneously in non-uniform circular motion.
• Forgetting that radians are dimensionless, preserving unit consistency.

Final Answer:a = α * r