Math  /  Algebra

QuestionAn object is travelling at a uniform speed in a circular path of radius 2.0 m . If the object's centripetal acceleration is found to be 3.0 m/s23.0 \mathrm{~m} / \mathrm{s}^{2}, what time is required for the object to make two complete rotations?

Studdy Solution

STEP 1

1. The object is traveling at a uniform speed in a circular path.
2. The radius of the circular path is 2.0m2.0 \, \text{m}.
3. The centripetal acceleration of the object is 3.0m/s23.0 \, \text{m/s}^2.
4. We need to find the time required for the object to make two complete rotations.

STEP 2

1. Use the formula for centripetal acceleration to find the speed of the object.
2. Calculate the circumference of the circular path.
3. Determine the time for one complete rotation.
4. Calculate the time for two complete rotations.

STEP 3

Use the formula for centripetal acceleration to find the speed of the object.
The formula for centripetal acceleration ac a_c is given by:
ac=v2r a_c = \frac{v^2}{r}
where v v is the speed of the object and r r is the radius of the circular path.
Given ac=3.0m/s2 a_c = 3.0 \, \text{m/s}^2 and r=2.0m r = 2.0 \, \text{m} , we can solve for v v :
3.0=v22.0 3.0 = \frac{v^2}{2.0}
Multiply both sides by 2.0 2.0 :
v2=6.0 v^2 = 6.0
Take the square root of both sides to solve for v v :
v=6.0 v = \sqrt{6.0}

STEP 4

Calculate the circumference of the circular path.
The circumference C C of a circle is given by:
C=2πr C = 2\pi r
Substitute r=2.0m r = 2.0 \, \text{m} :
C=2π×2.0=4πm C = 2\pi \times 2.0 = 4\pi \, \text{m}

STEP 5

Determine the time for one complete rotation.
The time T T for one complete rotation is given by:
T=Cv T = \frac{C}{v}
Substitute C=4πm C = 4\pi \, \text{m} and v=6.0 v = \sqrt{6.0} :
T=4π6.0 T = \frac{4\pi}{\sqrt{6.0}}

STEP 6

Calculate the time for two complete rotations.
The time for two complete rotations is:
2T=2×4π6.0 2T = 2 \times \frac{4\pi}{\sqrt{6.0}}
2T=8π6.0 2T = \frac{8\pi}{\sqrt{6.0}}
The time required for the object to make two complete rotations is:
8π6.0seconds \boxed{\frac{8\pi}{\sqrt{6.0}} \, \text{seconds}}

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