👤

It takes a time t = 0.64 s for a record to revolve on a record player once. A particular point on the record moves past the needle at a speed of vn = 1.02 m/s. what is the raduis at which the needle is incontact with the record?

Answer :

Givens.

• Time = 0.64 s.

,

• Tangential speed = 1.02 m/s.

The period is 0.64 seconds per revolution.

Then, we use the following formula.

[tex]v=\frac{2\pi}{T}\cdot r[/tex]

Where v = 1.02 m/s and T = 0.64 seconds. Solve for r.

[tex]\begin{gathered} r=\frac{v\cdot T}{2\pi} \\ r=\frac{1.02\cdot\frac{m}{s}\cdot0.64s}{2\pi} \\ r=\frac{0.6528}{2\pi}m \\ r\approx0.104m=10.4\operatorname{cm} \end{gathered}[/tex]

Therefore, the radius is 10.4 cm.

If the radius were 0.05 meters but using the same period, the tangential speed would be

[tex]\begin{gathered} v=\frac{2\pi}{T}\cdot r \\ v=\frac{2\pi}{0.64\sec}\cdot0.05m \\ v=0.50\cdot\frac{m}{s} \end{gathered}[/tex]

The tangential speed would be 0.50 meters per second, which is faster than the tangential speed with a larger radius.