In a loop-the-loop ride a car goes around a vertical, circular loop at a constan

Question

In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 273 kg and moves with speed v = 16.75 m/s. The loop-the-loop has a radius of R = 10.9 m.

1.What is the magnitude of the normal force on the care when it is at the bottom of the circle? (But as the car is accelerating upward.

2.What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward)?

3.What is the magnitude of the normal force on the car when it is at the top of the circle?

4.Compare the magnitude of the cars acceleration at each of the above locations:

abottom = aside = atop

abottom < aside < atop

abottom > aside > atop

5.What is the minimum speed of the car so that it stays in contact with the track at the top of the loop?

Explanation / Answer

Here ,

1)

at the bottom of the loop ,

Normal force = mg + mv^2/r

Normal force =272 *(9.8 + 16.75^2/10.9)

Normal force = 9702 N

the normal force at the bottom of circle is 9702 N

2)

at the side of circle

Normal force = m*v^2/R

Normal force = 273 * 16.75^2/10.9

Normal force = 7026.9 N

the normal force at the side of circle is 7026.9 N

3)

at the top of circle ,

Normal force = mv^2/R - mg

Normal force = 273 *(16.75^2/10.9 - 9.8)

Normal force = 4351.5 N

normal force at the top of the circle is 4351.5 N

4) as the speed of the car is constant ,

abottom = aside = atop

5)

let the minimum speed is v

mv^2/R = mg

v^2/10.9 = 9.8

v =10.34 m/s

the minimum speed at the top is 10.34 m/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.