Suppose you discover your grandfather's collection of Caruso records from the early part of the century. These disks played at 78.0 revolutions/minute (rpm) and were 9.00in. in diameter.
What was the radial acceleration of a point on the rim of such a record as you listened to the great tenor singing?
a=rω²
&omega=angular velocity in rad/s
r=radius, in m
a=acceleration in m/s²
7.63
To find the radial acceleration of a point on the rim of the record, we can use the formula for radial acceleration:
radial acceleration = (angular velocity)^2 * radius
First, we need to convert the units. We know that the record played at 78.0 revolutions per minute (rpm), which can be converted to radians per second (rad/s). Since one revolution is equal to 2π radians, we have:
angular velocity = (78.0 rpm) * (2π rad/1 rev) * (1 min/60 s)
= (78.0 rpm) * (2π rad/1 rev) * (1/60 s)
= 8.20 rad/s
Next, we need to convert the diameter of the record to its radius. The diameter is given as 9.00 inches, so the radius will be half of that:
radius = (9.00 in) / 2
= 4.50 in
Finally, we need to convert the radius to meters since the SI unit of acceleration is m/s^2. Since 1 inch is equal to 0.0254 meters, we have:
radius = (4.50 in) * (0.0254 m/1 in)
= 0.1143 m
Now, we can substitute the values into the formula to get the radial acceleration:
radial acceleration = (8.20 rad/s)^2 * 0.1143 m
= (67.24 rad^2/s^2) * 0.1143 m
= 7.68 m/s^2
Therefore, the radial acceleration of a point on the rim of the record as you listened to Caruso singing is approximately 7.68 m/s^2.