Question

6. An object is spun around in a circle of radius 3.0 m with speed of 3.77 m/s. What is its acceleration?

@1.25 m/s2 b. 4.74 m/s2 c. 0.79 m/s2 d. 2.39 m/s2
7. A 5.0 kg object is spun around in a circle of radius 2.5 m with speed of 4.5 m/s. What is its
centripetal force?
fa 9.0N
b. 2.8 N
c. 40.5 N
d. 45 N

Answer 1

6. Acceleration = 4.74 m/s^2

7. Centripetal force = 40.5 N

Step-by-step explanation:

Problem 6.

Recall that the centripetal acceleration is defined as:
`a_c=(v^2)/(r)`, where V is the object's tangential velocity, and r the radius of the circular motion. Therefore, in or case, the centripetal acceleration would be:

`a_c=(v^2)/(r)\\a_c=(3.77^2)/(3)\,(m)/(s^2) \\a_c=4.7376 (m)/(s^2)`

which we can round to 4.74 m/s^2 (option b in your list)

Problem 7.

Now we need to find not just the centripetal acceleration using the same formula as above, but then the centripetal force.

`a_c=(v^2)/(r)\\a_c=(4.5^2)/(2.5)\,(m)/(s^2) \\a_c=8.1 (m)/(s^2)`

Now we calculate the centripetal force by multiplying this acceleration times the mass of the object following the definition of force as mass times acceleration:

Centripetal force = 5.0 kg * 8.1 m/s^2 = 40.5 N

The answers comes in Newtons (N)