Question3. A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m . At point $A$ the speed of the car is $10.0 \mathrm{~m} / \mathrm{s}$ , and at point $B$ , the speed is $10.5 \mathrm{~m} / \mathrm{s}$ . Assume the child is not holding on and does not wear a seat belt. (a) What is the force of the car seat on the child at point $A$ ? ( $b$ ) What is the force of the car seat on the child at point $B$ ? (c) What minimum speed is required to keep the child in his seat at point $A$ ?

Studdy Solution

STEP 1

1. The child has a mass of $40.0 \, \text{kg}$ .
2. The roller coaster loop has a radius of $7.00 \, \text{m}$ .
3. The speed at point $A$ is $10.0 \, \text{m/s}$ .
4. The speed at point $B$ is $10.5 \, \text{m/s}$ .
5. Gravitational acceleration is $9.81 \, \text{m/s}^2$ .

STEP 2

1. Calculate the centripetal force at point $A$ .
2. Calculate the force of the car seat on the child at point $A$ .
3. Calculate the centripetal force at point $B$ .
4. Calculate the force of the car seat on the child at point $B$ .
5. Determine the minimum speed required to keep the child in the seat at point $A$ .

STEP 3

Calculate the centripetal force at point $A$ :
The formula for centripetal force is:
$F_c = \frac{m v^2}{r}$
Substitute the given values:
$F_c = \frac{40.0 \, \text{kg} \times (10.0 \, \text{m/s})^2}{7.00 \, \text{m}}$
$F_c = \frac{40.0 \times 100}{7.00}$
$F_c = \frac{4000}{7.00}$
$F_c = 571.43 \, \text{N}$

STEP 4

Calculate the force of the car seat on the child at point $A$ :
The force of the car seat is the difference between the gravitational force and the centripetal force:
$F_{\text{seat}} = F_c - mg$
$F_{\text{seat}} = 571.43 \, \text{N} - (40.0 \, \text{kg} \times 9.81 \, \text{m/s}^2)$
$F_{\text{seat}} = 571.43 \, \text{N} - 392.4 \, \text{N}$
$F_{\text{seat}} = 179.03 \, \text{N}$

STEP 5

Calculate the centripetal force at point $B$ :
$F_c = \frac{m v^2}{r}$
Substitute the given values:
$F_c = \frac{40.0 \, \text{kg} \times (10.5 \, \text{m/s})^2}{7.00 \, \text{m}}$
$F_c = \frac{40.0 \times 110.25}{7.00}$
$F_c = \frac{4410}{7.00}$
$F_c = 630.00 \, \text{N}$

STEP 6

Calculate the force of the car seat on the child at point $B$ :
$F_{\text{seat}} = F_c - mg$
$F_{\text{seat}} = 630.00 \, \text{N} - 392.4 \, \text{N}$
$F_{\text{seat}} = 237.6 \, \text{N}$

STEP 7

Determine the minimum speed required to keep the child in the seat at point $A$ :
At the top of the loop, the minimum speed occurs when the centripetal force equals the gravitational force:
$\frac{m v^2}{r} = mg$
$v^2 = rg$
$v = \sqrt{rg}$
Substitute the given values:
$v = \sqrt{7.00 \, \text{m} \times 9.81 \, \text{m/s}^2}$
$v = \sqrt{68.67}$
$v = 8.29 \, \text{m/s}$
The forces of the car seat on the child are:
(a) At point $A$ : $\boxed{179.03 \, \text{N}}$
(b) At point $B$ : $\boxed{237.6 \, \text{N}}$
(c) Minimum speed at point $A$ : $\boxed{8.29 \, \text{m/s}}$

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