A quarter on a spinning vinyl record has a centripetal acceleration of 34.0 m/s2. What is the acceleration of the quarter if it is placed twice as far from the centre of the vinyl record?
Ac = w^2 R
so doubling R fives twice the Ac, 68.0
A quarter on a spinning vinyl record has a centripetal acceleration of 17.3 m/s2. What is the acceleration of the quarter if it is placed twice as far from the centre of the vinyl record?
To find the acceleration of the quarter if it is placed twice as far from the center of the vinyl record, we can use the following formula:
a = ω^2 * r
Where:
a is the centripetal acceleration
ω is the angular velocity
r is the distance from the center of rotation
Given:
a = 34.0 m/s^2 (centripetal acceleration)
r = distance from the center of rotation (initial position)
Since the quarter is initially placed at a certain distance from the center, let's call it r1. Therefore, r1 is the distance from the center of rotation.
Since the quarter is then placed twice as far from the center, the new distance can be calculated as r2 = 2 * r1.
We know that the angular velocity (ω) remains constant as the vinyl record spins.
We need to find the new acceleration, a2, when the quarter is placed twice as far from the center. Using the formula, we have:
a1 = ω^2 * r1
Now, let's substitute r2 = 2 * r1 into the formula:
a2 = ω^2 * r2
= ω^2 * (2 * r1)
To find the relationship between a1 and a2, we can divide a1 by a2:
a1/a2 = (ω^2 * r1) / (ω^2 * (2 * r1))
= r1 / (2 * r1)
= 1/2
Therefore, a2 = (1/2) * a1
Substituting the given value of a1:
a2 = (1/2) * 34.0 m/s^2
= 17.0 m/s^2
So, the acceleration of the quarter when it is placed twice as far from the center of the vinyl record is 17.0 m/s^2.
To find the acceleration of the quarter when it is placed twice as far from the center of the vinyl record, we need to use the formula for centripetal acceleration:
a = (v^2) / r
where:
a = centripetal acceleration
v = tangential velocity
r = radius
We are given the initial centripetal acceleration, a = 34.0 m/s^2. We need to find the new acceleration when the quarter is placed twice as far from the center.
Let's call the initial radius r1 and the new radius r2. Since the new radius is twice as far, we can write:
r2 = 2 * r1
Now, to find the new acceleration, we need to find the new tangential velocity (v2). The tangential velocity is related to the radius and centripetal acceleration by the formula:
v = √(a * r)
Substituting the values:
v1 = √(34.0 m/s^2 * r1)
To find v2, we need to find the new radius r2. Using the relationship we found earlier: r2 = 2 * r1, we can substitute it into the formula for v2:
v2 = √(34.0 m/s^2 * (2 * r1))
Now we can find the new acceleration a2 using the formula:
a2 = (v2^2) / r2
Substituting the values we found for v2 and r2:
a2 = (√(34.0 m/s^2 * (2 * r1)))^2 / (2 * r1)
Simplifying the equation:
a2 = (34.0 m/s^2 * 2 * r1) / (2 * r1)
a2 = 34.0 m/s^2
Therefore, the acceleration of the quarter when it is placed twice as far from the center is still 34.0 m/s^2.