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In “Set 1: Linear Motion & Tangential Relationships” using v = rω, a point on the edge of a record player is 0.15 m from the center and rotates at 33 rad/s. What is the linear speed of that point?

Set 1: Linear Motion & Tangential Relationships Use s = rθ, v = rω, and aₜ = rα to solve these.
Set 1: Linear Motion & Tangential Relationships
Use s = rθ, v = rω, and aₜ = rα to solve these.
In “Set 1: Linear Motion & Tangential Relationships” using v = rω, a point on the edge of a record p...
Answer

Use the tangential speed relation $v=r\omega$. With $r=0.15\,\text{m}$ and $\omega=33\,\text{rad/s}$, $$v=(0.15)(33)=4.95\,\text{m/s}.$$ So the point’s linear speed is $4.95\,\text{m/s}$ (about $5.0\,\text{m/s}$).

Explanation

What you are converting here

The record player’s angular speed $\omega$ (how fast it turns in rad/s) can be turned into a linear or tangential speed $v$ (how fast a point on the rim moves in m/s) if you know the radius $r$.

Apply the tangential speed formula

For circular motion, $$v=r\omega.$$ Substitute the given values: $$v=(0.15\,\text{m})(33\,\text{rad/s})=4.95\,\text{m/s}.$$

Quick units check

Radians are dimensionless, so $(\text{m})(\text{rad/s})$ simplifies to $\text{m/s}$, which matches the units of linear speed.

Final result

$$v=4.95\,\text{m/s}.$$

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Skills You Achive
circular motion unit conversions algebraic substitution kinematics

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