A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its center, as shown in the figure below. The linear speed of a passenger on the rim is constant and equal to 6.00 m/s.
A.What is the magnitude of the passenger's acceleration as she passes through the lowest point in her circular motion?
B. What is the magnitude of the passenger's acceleration as she passes through the highest point in her circular motion?
C. How much time does it take the Ferris wheel to make one revolution?
A. We know that the linear speed (v) of the passenger on the rim is constant and equal to 6.00 m/s, and the radius (r) of the Ferris wheel is 14.0 m. The centripetal acceleration (a_c) at any point in the circular motion can be calculated using the formula:
a_c = v^2 / r
Plugging in the values, we get:
a_c = (6.00 m/s)^2 / 14.0 m
a_c = 36.00 m^2/s^2 / 14.0 m
a_c = 2.57 m/s^2
So, the magnitude of the passenger's acceleration as she passes through the lowest point in her circular motion is 2.57 m/s^2.
B. The magnitude of the centripetal acceleration is the same at any point in the circular motion. Therefore, the magnitude of the passenger's acceleration as she passes through the highest point in her circular motion is also 2.57 m/s^2.
C. To find the time it takes the Ferris wheel to make one revolution, we can use the formula:
t = 2πr/v
Plugging in the values, we get:
t = 2π(14.0 m) / 6.00 m/s
t = 28π m / 6.00 m/s
t ≈ 14.67 s
So, it takes the Ferris wheel approximately 14.67 seconds to make one revolution.
To answer these questions, we can use the equations of circular motion.
A. The acceleration of an object in circular motion can be calculated using the formula:
a = v^2 / r
where a is the acceleration, v is the linear speed, and r is the radius of the circular path.
Plugging in the given values, the acceleration at the lowest point is:
a = (6.00 m/s)^2 / 14.0 m
a = 36.0 m^2/s^2 / 14.0 m
a ≈ 2.57 m/s^2
Therefore, the magnitude of the passenger's acceleration as she passes through the lowest point is approximately 2.57 m/s^2.
B. Similarly, to calculate the acceleration at the highest point, we use the same formula:
a = v^2 / r
Plugging in the given values, the acceleration at the highest point is:
a = (6.00 m/s)^2 / 14.0 m
a = 36.0 m^2/s^2 / 14.0 m
a ≈ 2.57 m/s^2
Thus, the magnitude of the passenger's acceleration as she passes through the highest point is also approximately 2.57 m/s^2.
C. The time it takes for the Ferris wheel to make one revolution can be calculated using the formula:
T = 2πr / v
where T is the time period, r is the radius, and v is the linear speed of the passenger.
Plugging in the given values, the time period is:
T = 2π(14.0 m) / 6.00 m/s
T = 2π(14.0 m) / 6.00 m/s
T ≈ 14.7 s
Therefore, it takes approximately 14.7 seconds for the Ferris wheel to make one revolution.
To answer these questions, we need to understand the physics concepts related to circular motion and the equations associated with it.
A. To find the magnitude of the passenger's acceleration at the lowest point, we can use the formula for centripetal acceleration:
\[a=\frac{{v^2}}{{r}}\]
where \(a\) is the acceleration, \(v\) is the linear speed, and \(r\) is the radius of the circle.
In this case, the linear speed of the passenger is given as 6.00 m/s, and the radius of the Ferris wheel is 14.0 m. Plugging in these values into the formula, we get:
\[a=\frac{{(6.00\, \mathrm{m/s})^2}}{{14.0\, \mathrm{m}}}\]
Evaluating this expression, the magnitude of the passenger's acceleration at the lowest point is approximately 1.83 m/s².
B. The magnitude of the passenger's acceleration at the highest point can be found using the same formula:
\[a=\frac{{v^2}}{{r}}\]
At the highest point, the linear speed is still 6.00 m/s, but the radius of the circle changes to the sum of the radius of the Ferris wheel and the passenger's height from the lowest point. If the passenger is at the highest point, the height is equal to the radius of the Ferris wheel. Thus, the radius becomes \(14.0\, \mathrm{m} + 14.0\, \mathrm{m}\).
Substituting these values into the formula, we get:
\[a=\frac{{(6.00\, \mathrm{m/s})^2}}{{28.0\, \mathrm{m}}}\]
Evaluating this expression, the magnitude of the passenger's acceleration at the highest point is approximately 0.64 m/s².
C. To find the time taken for the Ferris wheel to make one revolution, we can use the equation for the period of circular motion:
\[T=\frac{{2\pi r}}{{v}}\]
where \(T\) is the period, \(r\) is the radius, and \(v\) is the linear speed.
In this case, the radius of the Ferris wheel is given as 14.0 m, and the linear speed is 6.00 m/s. Plugging in these values into the formula, we get:
\[T=\frac{{2\pi \times 14.0\, \mathrm{m}}}{{6.00\, \mathrm{m/s}}}\]
Evaluating this expression, the time taken for the Ferris wheel to make one revolution is approximately 14.7 seconds.