Question 1

A 1000-kg car is slowly picking up speed as it goes around a horizontal unbanked curve whose radius is 100 m. The coefficient of static friction between the tires and the road is 0.35. At what speed will the car begin to skid sideways?

Accepted Answer

The car will begin to skid sideways when the centripetal force required to keep the car moving in a circle exceeds the maximum static friction force. The maximum static friction force can be calculated using $f_{\text{max}} = \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. For a car on a horizontal surface, the normal force equals the weight of the car, $N = mg$, where $m$ is the mass of the car and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$).The centripetal force required for circular motion is given by $F_c = \frac{mv^2}{r}$, where $v$ is the velocity of the car and $r$ is the radius of the curve.Setting the maximum static friction force equal to the required centripetal force gives:$$\mu_s mg = \frac{mv^2}{r}$$Solving for $v$, we get:$$v = \sqrt{\mu_s g r}$$Substituting the given values ($\mu_s = 0.35$, $g = 9.8 \, \text{m/s}^2$, and $r = 100 \, \text{m}$):$$v = \sqrt{0.35 \times 9.8 \times 100} \approx 19 \, \text{m/s}$$Therefore, the car will begin to skid sideways at a speed of approximately 19 m/s.

Question 2

A Ferris wheel has radius 5.0 m and makes one revolution every 8.0 s with uniform rotation. A person who normally weighs 670 N is sitting on one of the benches attached at the rim of the wheel. What is the apparent weight (the normal force exerted on her by the bench)of the person as she passes through the highest point of her motion?

Accepted Answer

The answer of A Ferris wheel has radius 5.0 m...

Question 3

A 2.0-kg ball is moving with a constant speed of 5.0 m/s in a horizontal circle whose diameter is 1.0 m. What is the magnitude of the net force on the ball?

Accepted Answer

The answer of A 2.0-kg ball is moving with a...

Question 4

A 20-g bead is attached to a light 120-cm-long string as shown in the figure. This bead moves in a horizontal circle with a constant speed of 1.5 m/s. What is the tension in the string if the angle α is measured to be 25°?

Accepted Answer

The answer of A 20-g bead is attached to a...

Question 5

A 0.50-kg toy is attached to the end of a 1.0-m very light string. The toy is whirled in a horizontal circular path on a frictionless tabletop. If the maximum tension that the string can withstand without breaking is 350 N. What is the maximum speed the mass can have without breaking the string?

Accepted Answer

The answer of A 0.50-kg toy is attached to the...

Question 6

The curved section of a horizontal highway is a circular unbanked arc of radius 740 m. If the coefficient of static friction between this roadway and typical tires is 0.40, what would be the maximum safe driving speed for this horizontal curved section of highway?

Accepted Answer

The maximum safe driving speed on a horizontal curved section of a highway can be calculated using the formula for centripetal force and the maximum frictional force available due to static friction. The formula for the maximum velocity ($v_{max}$) in a circular turn on a level road is given by $v_{max} = \sqrt{\mu_s \cdot g \cdot r}$, where $\mu_s$ is the coefficient of static friction, $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$), and $r$ is the radius of the curve. Plugging in the given values: $v_{max} = \sqrt{0.40 \cdot 9.8 \cdot 740}$, we find that $v_{max} \approx 54 m/s$.

Question 7

Suppose our sun had 4 times its present mass but the earth orbited it at the same distance as it presently does. What would be the length of the year on the earth under those conditions?

Accepted Answer

The answer of Suppose our sun had 4 times its...

Question 8

A car moving at a steady 10 m/s on a level highway encounters a depression that has a circular cross-section with a radius of 30 m. The car maintains its speed as it drives through the depression. What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the bottom of the depression?

Accepted Answer

The normal force exerted by the seat of the car on the passenger can be found using the formula Fnet = ma, where Fnet is the net force acting on the passenger, m is the mass of the passenger, and a is the acceleration of the passenger. Since the car is maintaining a constant speed, the acceleration of the passenger is zero. Therefore, the net force acting on the passenger is zero, and the normal force exerted by the seat of the car on the passenger is equal in magnitude and opposite in direction to the force of gravity acting on the passenger. Using the formula Fg = mg, where g is the acceleration due to gravity (9.8 m/s^2), we can find the force of gravity acting on the passenger:

Fg = mg = (60.0 kg)(9.8 m/s^2) = 588 N

At the bottom of the depression, the passenger is experiencing an additional force due to the circular motion of the car. This force is called the centripetal force and is given by the formula Fc = mv^2/r, where v is the speed of the car and r is the radius of the depression.

Fc = mv^2/r = (60.0 kg)(10 m/s)^2/(30 m) = 200 N

The direction of the centripetal force is towards the center of the depression, so the normal force exerted by the seat of the car on the passenger must be greater than the force of gravity by an amount equal to the centripetal force:

Fn = Fg + Fc = 588 N + 200 N = 788 N

Therefore, the normal force exerted by the seat of the car on the passenger is 790 N (Choice C).

Question 9

A car moving at a steady 10 m/s on a level highway encounters a bump that has a circular cross-section with a radius of 30 m. The car maintains its speed over the bump. What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the top of the bump?

Accepted Answer

At the top of the bump, the passenger is experiencing maximum apparent weight due to the normal force, which is equal to the sum of actual weight and the centripetal force required for circular motion. 
Using Newton's second law, 
Net force = mass x acceleration 
The only force acting in the vertical direction is the normal force. Therefore, 
Normal force - weight =  mv²/r 
where m is the mass of the passenger, v is the speed of the car, and r is the radius of the bump. 
Substituting the given values, 
Normal force - (60.0 kg)(9.81 m/s²) = (60.0 kg)(10 m/s)²/(30 m) 
Normal force = 390 N

Question 10

A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the friction force required to keep the car from sliding?

Accepted Answer

The friction force required to keep the car from sliding can be calculated using the formula for centripetal force, $F = \frac{mv^2}{r}$, where $m$ is the mass of the car, $v$ is the velocity, and $r$ is the radius of the curve. The diameter of the curve is given as 0.20 km, which is 200 meters, so the radius $r$ is half of that, 100 meters. Plugging in the values, $F = \frac{(1000 \, \text{kg})(30 \, \text{m/s})^2}{100 \, \text{m}} = 9000 \, \text{N}$.

Question 11

One way that future space stations may create artificial gravity is by rotating the station. Consider a cylindrical space station 380 m in diameter that is rotating about its longitudinal axis. Astronauts walk on the inside surface of the space station. How long will it take for each rotation of the cylinder if it is to provide "normal" gravity for the astronauts?

Accepted Answer

To provide "normal" gravity through centripetal acceleration, the space station must simulate Earth's gravity (9.8 m/s^2). The formula for centripetal acceleration is $a = \frac{v^2}{r}$, where $v$ is the tangential velocity and $r$ is the radius of the circle. Since we're dealing with a cylindrical space station, we can also use the formula $a = \omega^2 \times r$, where $\omega$ is the angular velocity in radians per second, and $r$ is the radius of the cylinder. To find the period ($T$), which is the time for one complete rotation, we use the relation $T = \frac{2\pi}{\omega}$.Given that the diameter of the space station is 380 m, the radius ($r$) is half of that, which is 190 m. Setting the centripetal acceleration equal to Earth's gravity gives us:$$9.8 = \omega^2 \times 190$$Solving for $\omega$ gives:$$\omega = \sqrt{\frac{9.8}{190}}$$Then, to find the period ($T$), we use:$$T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{9.8}{190}}}$$Plugging in the numbers:$$T \approx 28 \text{ seconds}$$Therefore, it will take approximately 28 seconds for each rotation of the cylinder to provide "normal" gravity for the astronauts.

Question 12

A 250-kg motorcycle goes around an unbanked turn of radius 13.7 m at a steady 96.5 km/h. What is the magnitude of the net force on the motorcycle?

Accepted Answer

First, we need to convert the velocity from km/h to m/s:

96.5 km/h = 26.8 m/s

Next, we need to find the centripetal force acting on the motorcycle:

Fc = mv^2 / r

where Fc is the centripetal force, m is the mass of the motorcycle, v is the velocity, and r is the radius of the turn.

Plugging in the given values:

Fc = (250 kg)(26.8 m/s)^2 / 13.7 m 
Fc = 1.31 × 10^4 N

Finally, we need to find the net force acting on the motorcycle. Since the motorcycle is moving at a steady speed around a turn, we know that the net force must be equal to the centripetal force (otherwise, the motorcycle would be accelerating).

Therefore, the magnitude of the net force on the motorcycle is 1.31 × 10^4 N, which is answer choice C.

Question 13

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of the dive, the speed of the airplane is 150 m/s. What is the apparent weight of the 70-kg pilot at that point?

Accepted Answer

The apparent weight of the pilot can be calculated using the formula for centripetal force, which is necessary to keep the pilot moving in a circle. This force is given by $F = \frac{mv^2}{r}$, where $m$ is the mass, $v$ is the velocity, and $r$ is the radius of the circle. At the bottom of the dive, the centripetal force acts upwards, adding to the normal force (the pilot's weight) to give the apparent weight. The pilot's actual weight is $mg = 70 \times 9.8 = 686$ N (approximately 690 N). The centripetal force is $\frac{70 \times 150^2}{600} = 2625$ N. The apparent weight is the sum of the actual weight and the centripetal force: $686 + 2625 = 3311$ N, which is approximately 3300 N.

Question 14

If Earth had twice its present mass but it orbited at the same distance from the sun as it does now, its orbital period would be

Accepted Answer

The orbital period of a planet around the sun is determined by the distance from the sun, not by its mass, according to Kepler's third law of planetary motion. Therefore, even if Earth had twice its present mass, as long as it orbits at the same distance from the sun, its orbital period would remain 1 year.

Question 15

Satellite A has twice the mass of satellite B, and moves at the same orbital distance from Earth as satellite B. Compare the speeds of the two satellites.

Accepted Answer

The orbital speed of a satellite is determined by the gravitational pull of the Earth and its distance from the Earth, not by its own mass. Therefore, if two satellites are at the same orbital distance from Earth, they will have the same speed regardless of their masses.

Question 16

In a carnival ride, passengers stand with their backs against the wall of a cylinder. The cylinder is set into rotation and the floor is lowered away from the passengers, but they remain stuck against the wall of the cylinder. For a cylinder with a 2.0-m radius, what is the minimum speed that the passengers can have so they do not fall if the coefficient of static friction between the passengers and the wall is 0.25?

Accepted Answer

The minimum speed can be found using the formula for centripetal force and the force of friction. The force keeping the passengers from falling is the force of friction, which is $f = \mu N$, where $\mu$ is the coefficient of static friction and $N$ is the normal force, which in this case is provided by the centripetal force required to keep the passengers moving in a circle, $F_c = \frac{mv^2}{r}$. Setting the force of friction equal to the gravitational force (since this is what needs to be overcome to prevent falling), we get $\mu \frac{mv^2}{r} = mg$, where $m$ is the mass of the passenger, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $v$ is the velocity. Solving for $v$, we find $v = \sqrt{\frac{gr}{\mu}}$. Substituting $g = 9.8$ m/s$^2$, $r = 2.0$ m, and $\mu = 0.25$, we get $v = \sqrt{\frac{(9.8)(2.0)}{0.25}} = \sqrt{78.4} \approx 8.9$ m/s. This calculation shows that the mass of the passengers cancels out, indicating that the minimum speed does not depend on the mass of the passengers.

Question 17

A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop is $13.2 \mathrm {~m} ,$ with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car.

Accepted Answer

The minimum speed at the top of the loop can be found using the formula $ v = \sqrt{g \cdot r} $, where $ g $ is the acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $) and $ r $ is the radius of the loop. Plugging in the values, $ v = \sqrt{9.8 \times 13.2} \approx 11.4 \, \text{m/s} $.

Question 18

A jet plane flying 600 m/s experiences an acceleration of 4.0 g when pulling out of a circular dive. What is the radius of curvature of the circular part of the path in which the plane is flying?

Accepted Answer

The answer of A jet plane flying 600 m/s experiences...

Question 19

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle. At the bottom of the dive, the speed of the airplane is 320 m/s. What is the smallest radius allowable for the vertical circle if the pilot's apparent weight is not to exceed 7.0 times his true weight?

Accepted Answer

The apparent weight of the pilot at the bottom of the dive is the sum of the gravitational force and the centripetal force required to keep the pilot moving in a circle. This can be expressed as $W_{apparent} = W_{true} + F_{centripetal}$. The centripetal force can be calculated using the formula $F_{centripetal} = \frac{mv^2}{r}$, where $m$ is the mass of the pilot, $v$ is the velocity of the airplane, and $r$ is the radius of the circle. Given that the apparent weight is not to exceed 7 times the true weight, we can set up the equation $7W_{true} = W_{true} + \frac{mv^2}{r}$. Simplifying this, we get $6W_{true} = \frac{mv^2}{r}$, which leads to $r = \frac{mv^2}{6W_{true}}$. Since $W_{true} = mg$, where $g$ is the acceleration due to gravity (9.8 m/s$^2$), we can substitute to get $r = \frac{v^2}{6g}$. Plugging in the given values ($v = 320$ m/s), we get $r = \frac{(320)^2}{6 \times 9.8}$, which calculates to approximately 1700 meters.

Question 20

In order to simulate weightlessness for astronauts in training, they are flown in a vertical circle. If the passengers are to experience weightlessness, how fast should an airplane be moving at the top of a vertical circle with a radius of 2.5 km?

Accepted Answer

The answer of In order to simulate weightlessness for astronauts...

Quiz 6: Gravity