Answer :
Answer:
Approximately [tex]77.1\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
If an object is in a circular motion of radius [tex]r[/tex] at an angular velocity of [tex]\omega[/tex], the centripetal acceleration of the object would be:
[tex]a = \omega^{2} \, r[/tex].
In this question, the angular velocity of the object is given in the unit [tex]\text{rpm}[/tex], or "revolutions per minute". Apply unit conversion and ensure that angular velocity is measured in the standard unit of (radians) per second:
- [tex]1\; {\text{minute}} = 60\; {\rm s}[/tex].
- [tex]1\; \text{revolution} = 2\, \pi[/tex] (radians.)
Hence:
[tex]\begin{aligned}\omega &= 75\; \text{rpm} \\ &= 75\; {\rm (\text{revolution}) \, (\text{minute})^{-1}} \times \frac{1\; \text{minute}}{60\; {\rm s}}\times \frac{2\,\pi}{1\; \text{revolution}} \\ &= \frac{5\, \pi}{2}\; {\rm s^{-1}}\end{aligned}[/tex].
The centripetal acceleration of this object would be:
[tex]\begin{aligned} a &= \omega^{2} \, r \\ &\approx \left(\frac{5\, \pi}{2}\; {\rm s^{-1}\right)^{2} \, (1.25\; {\rm m}) \\ & \approx 77.1\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
The correct option c. 77.1 m/s. is The magnitude of the centripetal acceleration is 77.1 m/s.
To determine the magnitude of the centripetal acceleration of a merry-go-round, we need to use the formula:
ac = rω²
where ac is the centripetal acceleration, r is the radius, and ω is the angular velocity in radians per second.
Step-by-step Solution:
First, convert the rotation speed from revolutions per minute (rpm) to radians per second (rad/s). The merry-go-round rotates at 75 rpm.
Convert rpm to radians per second:
1 rpm = 2π radians / 60 seconds
ω = 75 rpm × (2π / 60) = 7.85 rad/s
Next, use the radius of the merry-go-round (r = 1.25 m) and the angular velocity in the centripetal acceleration formula:
Calculate the centripetal acceleration:
ac = rω²
ac = 1.25 m × (7.85 rad/s)²
ac ≈ 77.1 m/s²
The correct answer is option c. 77.1 m/s².
