Answer:
If the radius of a particle rotating in a circle is doubled without changing the particle's speed, the centripetal acceleration (a) will decrease by a factor of 4.
To understand why, let's look at the formula for centripetal acceleration:
a = v^2 / r
where v is the speed of the particle and r is the radius of the circle. Since the speed is not changing, if we double the radius (r), the centripetal acceleration will decrease by a factor of (2^2) = 4.
So, if the initial centripetal acceleration is 7.0 m/s^2, when the radius is doubled, the new centripetal acceleration would be 7.0 m/s^2 / 4 = 1.75 m/s^2.
Now, let's consider the scenario where the speed is doubled without changing the circle's radius. In this case, the centripetal acceleration will increase by a factor of 4.
Using the same formula, we can rearrange it to solve for v:
v = sqrt(a * r)
If the speed is doubled, it means the new speed (v') is 2 times the initial speed (v). Plugging this into the formula, we get:
2v = sqrt(a * r)
Squaring both sides of the equation, we have:
(2v)^2 = a * r
4v^2 = a * r
Since r is not changing, if we double the speed (v), the centripetal acceleration (a) will increase by a factor of 4.
Therefore, if the initial centripetal acceleration is 7.0 m/s^2, when the speed is doubled, the new centripetal acceleration would be 7.0 m/s^2 * 4 = 28.0 m/s^2.
Explanation:
Answer:
If the radius is doubled without changing the particle's speed, the centripetal acceleration becomes half the original value.
If the speed is doubled without changing the circle's radius, the centripetal acceleration becomes four times the original value.
Explanation:
To solve these problems, we can use the centripetal acceleration formula:
[tex] \Large\boxed{\boxed{a = \dfrac{v^2}{r}}} [/tex]
Where:
Let's address each scenario:
If the radius is doubled without changing the particle's speed:
Let [tex] a_1 [/tex] be the initial acceleration and [tex] a_2 [/tex] be the new acceleration.
Given that the radius is doubled ([tex] r_2 = 2r_1 [/tex]) and the speed remains the same, we can use the centripetal acceleration formula to compare the initial and new situations:
[tex] a_1 = \dfrac{v^2}{r_1} [/tex]
[tex] a_2 = \dfrac{v^2}{2r_1} [/tex]
Now, we see that [tex] a_2 [/tex] is half of [tex] a_1 [/tex] because [tex] r_2 = 2r_1 [/tex].
So, if the radius is doubled without changing the particle's speed, the centripetal acceleration becomes half the original value.
[tex]\hrulefill[/tex]
If the speed is doubled without changing the circle's radius:
Let [tex] a_1 [/tex] be the initial acceleration and [tex] a_2 [/tex] be the new acceleration.
Given that the speed is doubled ([tex] v_2 = 2v_1 [/tex]) and the radius remains the same, we can use the centripetal acceleration formula to compare the initial and new situations:
[tex] a_1 = \dfrac{v_1^2}{r} [/tex]
[tex] a_2 = \dfrac{v_2^2}{r} [/tex]
Now, if the speed is doubled, the new acceleration becomes four times the original acceleration because [tex] v_2 = 2v_1 [/tex]:
[tex] a_2 = \dfrac{(2v_1)^2}{r} \\\\= \dfrac{4v_1^2}{r} \\\\= 4a_1 [/tex]
So, if the speed is doubled without changing the circle's radius, the centripetal acceleration becomes four times the original value.
