Question 1

A 45-kg woman is riding a Ferris wheel of radius 15 m at a constant speed.The speed of the Ferris wheel that would make her weightless at the top is

Accepted Answer

At the top of the Ferris wheel, for the woman to feel weightless, the only force acting on her must be the centripetal force, which is equal to the gravitational force at that point. The formula for centripetal force is $F_c = \frac{mv^2}{r}$, where $m$ is the mass, $v$ is the velocity, and $r$ is the radius. The gravitational force is $F_g = mg$, where $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$). Setting $F_c = F_g$, we get $\frac{mv^2}{r} = mg$. Simplifying, $v^2 = rg$, and solving for $v$, we get $v = \sqrt{rg}$. Substituting $r = 15 m$ and $g = 9.8 m/s^2$, we find $v = \sqrt{15 \times 9.8} \approx 12.1 m/s$.

Question 2

A CD-ROM starts from rest and undergoes uniform angular acceleration until it reaches 26,000 rpm in 0.25 s.What is its angular acceleration?

Accepted Answer

The angular acceleration is given by:$$\alpha = \frac{\omega_f - \omega_i}{t}$$where:* $\omega_f$ is the final angular velocity* $\omega_i$ is the initial angular velocity* $t$ is the timeIn this case, $\omega_i = 0$, so:$$\alpha = \frac{\omega_f}{t} = \frac{26,000 \text{ rpm}}{0.25 \text{ s}} = 1.1 \times 10^4 \text{ rad/s}^2$$

Question 3

How far does a person in Dubuque,IA (latitude 42.5S1U1P1$\circ$S1S1P0)move in the 7.50 hours she sleeps? The radius of the Earth is 6378 km.

Accepted Answer

We can use the formula:

distance = (angular speed) x (radius of Earth) x (time)

The angular speed for a person at rest on the Earth's surface is 0 degrees per hour. So, we can calculate the distance as:

distance = (0 degrees/hour) x (6378 km) x (7.5 hours) x cos(42.5 degrees)

Using a calculator, this simplifies to approximately 9230 km. Therefore, the best choice is C.

Question 4

The surface of the exit road is horizontal,not banked.(See figure.)If the static friction between the tires and the surface of the road is µs = 0.688 and the maximum speed with which the car can exit the highway safely without sliding is 25.2 m/s,what is the radius of curvature of a highway exit ,r?

Accepted Answer

The answer of The surface of the exit road is...

Question 5

A small object with a mass of m = 658 g is whirled at the end of a rope in a vertical circle with a radius of r = 137 cm.When it is at the location shown,(mid-height),and its speed is v = 5.54 m/s,calculate the direction of the total force acting on the mass.(as measured from a line drawn from m to the center of the circular path)

Accepted Answer

The answer of A small object with a mass of...

Question 6

An orbital space station generates "artificial" gravity by spinning.Consider a 100-kg person standing against the outside wall of a doughnut shaped space station with an outer diameter of 2.0 km.What is the period of rotation for the space station such that the space station exerts a force on the person equal to the weight of the person on the surface of the Earth?

Accepted Answer

The answer of An orbital space station generates "artificial" gravity...

Question 7

A 2.0-m long string is tied to a 3.0-kg rock and the rock is twirled around in a circle at a constant speed.The speed of the rock is such that the string makes an angle of 45 degrees with the horizontal.How long does it take the rock to complete one full circle?

Accepted Answer

The tension in the string provides the centripetal force and a component to balance the weight of the rock. Using trigonometry and the centripetal force formula, we can find the speed and then the period of rotation. The calculations show that the time to complete one full circle is approximately 2.4 seconds.

Question 8

A 45-kg woman is riding a Ferris wheel of radius 15 m at a constant speed.The angular speed of the Ferris wheel that would make her weightless at the top is

Accepted Answer

For a person to feel weightless at the top of the Ferris wheel, the centripetal force must equal the gravitational force acting on them. The centripetal force ($F_c$) is given by $F_c = m\omega^2r$, where $m$ is the mass, $\omega$ is the angular speed, and $r$ is the radius. The gravitational force ($F_g$) is given by $F_g = mg$, where $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$). Setting $F_c = F_g$, we get $m\omega^2r = mg$. Simplifying, $\omega^2 = g/r$. Substituting $g = 9.8 m/s^2$ and $r = 15 m$, we find $\omega = \sqrt{9.8/15}$, which calculates to approximately $0.81 rad/s$.

Question 9

A car with a mass of 1200 kg goes over a hill at a constant speed.The top of the hill can be approximated as an arc length of a circle with a radius of curvature of 210 m.What minimum speed must the car have as it passes over the top of the hill in order to become airborn?

Accepted Answer

The minimum speed required for the car to become airborne at the top of the hill is the speed at which the normal force acting on the car is zero. At this point, the only forces acting on the car are gravity and the force of the road pushing up on the car. The normal force is given by:$$N = mg - mv^2/r$$where m is the mass of the car, g is the acceleration due to gravity, v is the speed of the car, and r is the radius of curvature of the hill.Setting N = 0, we get:$$mg - mv^2/r = 0$$Solving for v, we get:$$v = sqrt(rg)$$Substituting the given values, we get:$$v = sqrt(210 m * 9.8 m/s^2) = 45 m/s$$Therefore, the minimum speed the car must have as it passes over the top of the hill in order to become airborne is 45 m/s.

Question 10

The radius of curvature of a highway exit is r = 68.5 m.The surface of the exit road is horizontal,not banked.(See figure.)If the static friction between the tires and the surface of the road is µs = 0.688,then what is the maximum speed at which the car can exit the highway safely without sliding? (in m/s)

Accepted Answer

The answer of The radius of curvature of a highway...

Question 11

A CD-ROM starts from rest and undergoes uniform angular acceleration until it reaches 26,000 rpm in 0.25 s.Through what angle does it rotate during this time?

Accepted Answer

First, we need to convert the final angular velocity to radians per second:

26,000 rpm = 26,000/60 revolutions per second = 433.33... revolutions per second 
Each revolution is 2π radians, so:

433.33... revolutions per second x 2π radians/revolution = 2,725.6 radians per second

The angular acceleration can be found using the equation:

ωf = ωi + αt

Where ωi = 0 (starting from rest), ωf = 2,725.6 radians per second, t = 0.25 seconds, and solving for α:

α = (ωf - ωi)/t = 10,902.4 radians per second squared

Finally, we can use the equation for rotational displacement:

θ = ωit + (1/2)αt^2

Where ωi = 0, t = 0.25 seconds, and solving for θ:

θ = (1/2)αt^2 = 340 radians

Therefore, the answer is C, 340 rad.

Question 12

A small object with a mass of m = 658 g is whirled at the end of a rope in a vertical circle with a radius of r = 137 cm.When it is at the location shown,(mid-height),and its speed is v = 5.54 m/s,calculate the magnitude of the total force acting on the mass.(in N)

Accepted Answer

The answer of A small object with a mass of...

Question 13

A car drives along a country road at a constant speed of 26 m/s.When it encounters a hill,its tires just barely leave the pavement as the car passes over the top of the hill.What is the radius of curvature of the hill?

Accepted Answer

The radius of curvature of the hill can be found using the formula for centripetal acceleration $a = \frac{v^2}{r}$, where $a$ is the acceleration due to gravity (9.8 m/s$^2$) when the car just barely leaves the ground, $v$ is the velocity (26 m/s), and $r$ is the radius of curvature. Rearranging the formula to solve for $r$, we get $r = \frac{v^2}{a}$. Substituting the given values, $r = \frac{26^2}{9.8} \approx 69$ meters.

Question 14

A small object with a mass of m = 658 g is whirled at the end of a rope in a vertical circle with a radius of r = 137 cm.When it is at the location shown,(mid-height),and its speed is v = 5.54 m/s.Determine the tension in the rope.(in N)

Accepted Answer

The answer of A small object with a mass of...

Question 15

A 2.0-m long string is tied to a 3.0-kg rock and the rock is twirled around in a circle at a constant speed.The speed of the rock is such that the string makes an angle of 45 degrees with the horizontal.What is the speed of the rock?

Accepted Answer

The tension in the string provides the centripetal force keeping the rock moving in a circle. The tension T in the string is given by T = (m*v^2)/r, where m is the mass of the rock, v is its velocity, and r is the radius of the circle (equal to the length of the string). Since the string makes an angle of 45 degrees with the horizontal, the vertical component of the tension balances the weight of the rock, and the horizontal component provides the centripetal force. This gives T*cos(45) = m*g, or T = m*g/cos(45), where g is the acceleration due to gravity. Substituting this expression for T into the equation for centripetal force gives (m*v^2)/r = T*sin(45), or v = sqrt((r*g*tan(45))/m) = sqrt((2*9.81*tan(45))/3) =~ 3.7 m/s. Therefore, the best choice is C.

Question 16

A child's top is spinning at a rate of 7.00 rev/s.It takes 30.0 seconds for the top to slow to a rate of 2.00 rev/s.Assuming a constant angular acceleration,what is the magnitude of the angular acceleration of the top?

Accepted Answer

The answer of A child's top is spinning at a...

Question 17

Suppose the people standing at rest with respect to the Earth on the Earth's equator are weightless.The radius of the earth is 6.37 * 10⁶ m.The length of the day under this condition would be

Accepted Answer

For people standing at rest with respect to the Earth on the Earth's equator to be weightless, the centripetal force required to keep them moving in a circle must equal the gravitational force. The formula for centripetal force is $F_c = \frac{mv^2}{r}$, and the gravitational force is given by $F_g = \frac{GmM}{r^2}$, where $m$ is the mass of the object (person), $v$ is the tangential velocity, $r$ is the radius of the Earth, $G$ is the gravitational constant, and $M$ is the mass of the Earth.Setting these two forces equal to each other and solving for $v$, we get $v = \sqrt{\frac{GM}{r}}$. The circumference of the Earth is $2\pi r$, so the time $T$ to make one rotation (the length of the day) is $T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r}{G M}}$.Plugging in the values for $G$, $M$, and $r$ (with $r = 6.37 \times 10^6 m$), and solving for $T$, we find that $T$ is approximately 84.4 minutes. This calculation assumes that the only force acting on the people is gravity (which is being counteracted by the centripetal force due to their motion), making them feel weightless.

Question 18

A CD-ROM starts from rest and undergoes uniform angular acceleration until it reaches 26,000 rpm in 0.25 s.If its radius is 0.061 m,how far does a point on its edge travel during this time?

Accepted Answer

First, we need to convert the final angular velocity from rpm to rad/s:
ω = (26000 rpm) x (2π rad/1 rev) x (1 min/60 s) x (1/60 s) = 2717.9 rad/s

Next, we can use the kinematic equation for angular motion:
θ = ω₀t + 1/2αt²
where θ is the total angle turned, ω₀ is the initial angular velocity (0 in this case), α is the angular acceleration, and t is the elapsed time.

To solve for θ, we need to find α first:
α = ω/ t = (2717.9 rad/s) / (0.25 s) = 10871.6 rad/s²

Now we can plug in the values to get θ:
θ = 1/2 (10871.6 rad/s²) (0.25 s)² = 678.5 rad

Finally, we can find the distance traveled by a point on the edge of the CD:
d = rθ = (0.061 m) (678.5 rad) = 41.4 m ≈ 21 m (rounded to one significant figure)

Question 19

The wheel of a racing car is rotating at 200 rpm at a given instant of time.In the next minute it makes 100 revolutions.The angular acceleration of the wheel is

Accepted Answer

The answer of The wheel of a racing car is...

Question 20

The surface of the exit road is horizontal,not banked.(See figure.)If the radius of curvature of a highway exit is 82.5 m and the maximum speed with which the car can exit the highway safely without sliding is 25.2 m/s,what is the coefficient of static friction between the tires and the surface of the road?

Accepted Answer

The maximum speed that the car can take the exit safely without sliding can be found using the formula:

vmax = sqrt(μrg)

where μ is the coefficient of static friction, r is the radius of curvature, and g is the acceleration due to gravity.

Plugging in the given values, we get:

25.2 = sqrt(μ x 9.8 x 82.5)

Squaring both sides and rearranging, we get:

μ = (25.2/9.8)^2/82.5 = 0.78

Therefore, the coefficient of static friction between the tires and the surface of the road is 0.78. The closest answer choice is C.

Quiz 9: Circular Motion

A 45-kg woman is riding a Ferris wheel of radius 15 m at a constant speed.The speed of the Ferris wheel that would make her weightless at the top is

A CD-ROM starts from rest and undergoes uniform angular acceleration until it reaches 26,000 rpm in 0.25 s.What is its angular acceleration?

How far does a person in Dubuque,IA (latitude 42.5^$\circ$) move in the 7.50 hours she sleeps? The radius of the Earth is 6378 km.

A 2.0-m long string is tied to a 3.0-kg rock and the rock is twirled around in a circle at a constant speed.The speed of the rock is such that the string makes an angle of 45 degrees with the horizontal.How long does it take the rock to complete one full circle?

A 45-kg woman is riding a Ferris wheel of radius 15 m at a constant speed.The angular speed of the Ferris wheel that would make her weightless at the top is

A car with a mass of 1200 kg goes over a hill at a constant speed.The top of the hill can be approximated as an arc length of a circle with a radius of curvature of 210 m.What minimum speed must the car have as it passes over the top of the hill in order to become airborn?

A CD-ROM starts from rest and undergoes uniform angular acceleration until it reaches 26,000 rpm in 0.25 s.Through what angle does it rotate during this time?

A small object with a mass of m = 658 g is whirled at the end of a rope in a vertical circle with a radius of r = 137 cm.When it is at the location shown,(mid-height) ,and its speed is v = 5.54 m/s,calculate the magnitude of the total force acting on the mass.(in N)

A car drives along a country road at a constant speed of 26 m/s.When it encounters a hill,its tires just barely leave the pavement as the car passes over the top of the hill.What is the radius of curvature of the hill?

A small object with a mass of m = 658 g is whirled at the end of a rope in a vertical circle with a radius of r = 137 cm.When it is at the location shown,(mid-height) ,and its speed is v = 5.54 m/s.Determine the tension in the rope.(in N)

A 2.0-m long string is tied to a 3.0-kg rock and the rock is twirled around in a circle at a constant speed.The speed of the rock is such that the string makes an angle of 45 degrees with the horizontal.What is the speed of the rock?

A child's top is spinning at a rate of 7.00 rev/s.It takes 30.0 seconds for the top to slow to a rate of 2.00 rev/s.Assuming a constant angular acceleration,what is the magnitude of the angular acceleration of the top?

Suppose the people standing at rest with respect to the Earth on the Earth's equator are weightless.The radius of the earth is 6.37 * 10⁶ m.The length of the day under this condition would be

A CD-ROM starts from rest and undergoes uniform angular acceleration until it reaches 26,000 rpm in 0.25 s.If its radius is 0.061 m,how far does a point on its edge travel during this time?

The wheel of a racing car is rotating at 200 rpm at a given instant of time.In the next minute it makes 100 revolutions.The angular acceleration of the wheel is

Quiz 9: Circular Motion

A 45-kg woman is riding a Ferris wheel of radius 15 m at a constant speed.The speed of the Ferris wheel that would make her weightless at the top is

A CD-ROM starts from rest and undergoes uniform angular acceleration until it reaches 26,000 rpm in 0.25 s.What is its angular acceleration?

How far does a person in Dubuque,IA (latitude 42.5 ∘\circ∘ ) move in the 7.50 hours she sleeps? The radius of the Earth is 6378 km.

A 2.0-m long string is tied to a 3.0-kg rock and the rock is twirled around in a circle at a constant speed.The speed of the rock is such that the string makes an angle of 45 degrees with the horizontal.How long does it take the rock to complete one full circle?

A 45-kg woman is riding a Ferris wheel of radius 15 m at a constant speed.The angular speed of the Ferris wheel that would make her weightless at the top is

A car with a mass of 1200 kg goes over a hill at a constant speed.The top of the hill can be approximated as an arc length of a circle with a radius of curvature of 210 m.What minimum speed must the car have as it passes over the top of the hill in order to become airborn?

A CD-ROM starts from rest and undergoes uniform angular acceleration until it reaches 26,000 rpm in 0.25 s.Through what angle does it rotate during this time?

A small object with a mass of m = 658 g is whirled at the end of a rope in a vertical circle with a radius of r = 137 cm.When it is at the location shown,(mid-height) ,and its speed is v = 5.54 m/s,calculate the magnitude of the total force acting on the mass.(in N)

A car drives along a country road at a constant speed of 26 m/s.When it encounters a hill,its tires just barely leave the pavement as the car passes over the top of the hill.What is the radius of curvature of the hill?

A small object with a mass of m = 658 g is whirled at the end of a rope in a vertical circle with a radius of r = 137 cm.When it is at the location shown,(mid-height) ,and its speed is v = 5.54 m/s.Determine the tension in the rope.(in N)

A 2.0-m long string is tied to a 3.0-kg rock and the rock is twirled around in a circle at a constant speed.The speed of the rock is such that the string makes an angle of 45 degrees with the horizontal.What is the speed of the rock?

A child's top is spinning at a rate of 7.00 rev/s.It takes 30.0 seconds for the top to slow to a rate of 2.00 rev/s.Assuming a constant angular acceleration,what is the magnitude of the angular acceleration of the top?

Suppose the people standing at rest with respect to the Earth on the Earth's equator are weightless.The radius of the earth is 6.37 * 106 m.The length of the day under this condition would be

A CD-ROM starts from rest and undergoes uniform angular acceleration until it reaches 26,000 rpm in 0.25 s.If its radius is 0.061 m,how far does a point on its edge travel during this time?

The wheel of a racing car is rotating at 200 rpm at a given instant of time.In the next minute it makes 100 revolutions.The angular acceleration of the wheel is

How far does a person in Dubuque,IA (latitude 42.5^$\circ$) move in the 7.50 hours she sleeps? The radius of the Earth is 6378 km.

Suppose the people standing at rest with respect to the Earth on the Earth's equator are weightless.The radius of the earth is 6.37 * 10⁶ m.The length of the day under this condition would be