Final answer:
The period of rotation for an LP with a radius of approximately 25 cm and centripetal acceleration of 1.87 m/s2 can be found by rearranging the centripetal acceleration formula in terms of period and solving for it using the given values.
Step-by-step explanation:
To find the period of rotation of the LP (long-playing phonograph record) given a centripetal acceleration and a radius, we use the relationship between the centripetal acceleration, radius, and the angular velocity. Centripetal acceleration (ac) can be defined using the following equation, where v is the linear velocity and r is the radius: ac = (v2) / r.
However, since we are looking for the period (T), which is the time for one complete rotation, we can rearrange the equation in terms of velocity and period: v = (2 π r) / T. By substituting the expression for v into the centripetal acceleration formula, we then have: ac = (4 π2 r) / T2. This allows us to solve for T using the given centripetal acceleration (1.87 m/s2) and radius (0.25 m).
Solving for T gives us T = sqrt(4 π2 r / ac). Plugging in the given values, we get T = sqrt(4 π2 * 0.25 m / 1.87 m/s2) which yields the period of rotation for the LP.