Question 1

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 2

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?
A)9.3 m/s
B)24 m/s
C)34 m/s
D)35 m/s
E)19 m/s

Accepted Answer

The speed at which the car will begin to skid sideways can be found using the formula for centripetal force required to keep the car moving in a circle, $F = \frac{mv^2}{r}$, where $m$ is the mass of the car, $v$ is the velocity, and $r$ is the radius of the curve. The maximum force of static friction, which prevents the car from skidding, is $F_{friction} = \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. On a flat surface, $N = mg$, where $g$ is the acceleration due to gravity (9.8 m/s$^2$). Setting the centripetal force equal to the maximum frictional force gives $\frac{mv^2}{r} = \mu_s mg$. Solving for $v$, we get $v = \sqrt{\mu_s g r}$. Substituting $\mu_s = 0.35$, $g = 9.8 m/s^2$, and $r = 100 m$, we find $v = \sqrt{0.35 \times 9.8 \times 100} = \sqrt{343} \approx 18.5 m/s$, which rounds to 19 m/s, making option E the correct answer.

Question 3

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?
A)54 m/s
B)52 m/s
C)50 m/s
D)48 m/s
E)46 m/s

Accepted Answer

The maximum safe driving speed on a curved section of a horizontal highway can be calculated using the formula for centripetal force and the maximum force of static friction. The formula for the maximum velocity ($v$) in a circular turn on a level road is given by $v = \sqrt{\mu_s \cdot g \cdot r}$, 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 ($\mu_s = 0.40$, $g = 9.8 m/s^2$, and $r = 740 m$), we get:$$v = \sqrt{0.40 \cdot 9.8 \cdot 740} \approx \sqrt{2929.6} \approx 54.13 m/s$$Therefore, the maximum safe driving speed is approximately 54 m/s.

Question 4

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 force exerted by the seat of the car on a 60.0-kg passenger when the car is at the top of the bump?
A)200 N
B)390 N
C)790 N
D)490 N
E)590 N

Accepted Answer

At the top of the bump, the forces acting on the passenger are the normal force (N) from the seat and the gravitational force (mg). The net force provides the centripetal force needed to move in a circle. Thus, $N + mg = \frac{mv^2}{r}$. Rearranging for N gives $N = \frac{mv^2}{r} - mg$. Substituting $m = 60.0 \, \text{kg}$, $v = 10 \, \text{m/s}$, $r = 30 \, \text{m}$, and $g = 9.8 \, \text{m/s}^2$, we get $N = \frac{60 \times 100}{30} - 60 \times 9.8 = 200 - 588 = -388 \, \text{N}$. The negative sign indicates the direction of the force, but we are interested in the magnitude, which is 388 N, rounded to 390 N for significant figures.

Question 5

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?
A)28 s
B)39 s
C)6.2 s
D)4.4 s

Accepted Answer

To provide "normal" gravity through centripetal acceleration, the formula $a = \frac{v^2}{r}$ can be used, where $a$ is the acceleration due to gravity on Earth ($9.8 m/s^2$), $v$ is the tangential velocity, and $r$ is the radius of the cylinder. However, for this problem, it's more straightforward to use the formula relating the rotational speed to the acceleration: $a = \omega^2 \times r$, where $\omega$ is the angular velocity in radians per second, and $r$ is the radius. Rearranging this formula to solve for $\omega$, we get $\omega = \sqrt{\frac{a}{r}}$. Given that the diameter is 380 m, the radius $r$ is 190 m. Plugging in the values, $\omega = \sqrt{\frac{9.8}{190}}$, which calculates to an angular velocity. To find the period $T$ (time for one rotation), use $T = \frac{2\pi}{\omega}$. Substituting the calculated $\omega$ gives the period $T$, which closely matches the option for 28 seconds. This approach uses physics principles of circular motion to determine the necessary rotation period to simulate Earth's gravity.

Question 6

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 (the force that the pilot exerts on the seat)is not to exceed 7.0 times his true weight?
A)1700 m
B)1500 m
C)2200 m
D)230 m
E)42 m

Accepted Answer

The pilot's apparent weight is the sum of his true weight and the centripetal force required to keep him moving in a circle. This can be expressed as $F_{apparent} = mg + \frac{mv^2}{r}$, where $m$ is the mass of the pilot, $g$ is the acceleration due to gravity (9.8 m/s$^2$), $v$ is the velocity of the airplane, and $r$ is the radius of the circle. Given that the apparent weight is 7 times the true weight, we can set up the equation $7mg = mg + \frac{mv^2}{r}$. Simplifying this gives $6mg = \frac{mv^2}{r}$, and solving for $r$ gives $r = \frac{v^2}{6g}$. Substituting $v = 320$ m/s and $g = 9.8$ m/s$^2$ gives $r = \frac{(320)^2}{6 \times 9.8} \approx 1700$ meters.

Question 7

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 sum of the forces exerted on the motorcycle? A)719 N B)2.95 × 10³ N C)1.31 × 10⁴ N D)4.31 × 10⁴ N

Accepted Answer

First, we need to find the centripetal acceleration acting on the motorcycle.

v = 96.5 km/h = 26.8 m/s (convert to m/s) 
r = 13.7 m

a = v^2/r = (26.8 m/s)^2 / 13.7 m = 52.4 m/s^2

Next, we can use Newton's second law to find the force (F) exerted on the motorcycle.

F = ma = (250 kg) x (52.4 m/s^2) = 13,100 N

Lastly, we need to consider any other forces acting on the motorcycle. We can assume that the normal force and the force due to friction are the only other forces present. Since the motorcycle is not banking, the normal force and force due to friction must add up to be equal in magnitude to the centripetal force.

Therefore, the magnitude of the sum of forces is 2F = 2(13,100 N) = 26,200 N or 1.31 × 10^14 N (in scientific notation).

The correct choice is C.

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 force exerted by the seat of the car on a 60.0-kg passenger when the car is at the bottom of the depression?
A)200 N
B)390 N
C)790 N
D)490 N
E)590 N

Accepted Answer

The force exerted by the seat of the car on the passenger is the normal force, which is equal to the weight of the passenger plus the centripetal force required to keep the passenger moving in a circle. The centripetal force is given by:$$F_c = mv^2/r$$where:* m is the mass of the passenger* v is the speed of the car* r is the radius of the circular pathIn this case, the mass of the passenger is 60.0 kg, the speed of the car is 10 m/s, and the radius of the circular path is 30 m. Therefore, the centripetal force is:$$F_c = (60.0 kg)(10 m/s)^2/30 m = 200 N$$The normal force is equal to the weight of the passenger plus the centripetal force, which is:$$F_n = mg + F_c = (60.0 kg)(9.8 m/s^2) + 200 N = 790 N$$Therefore, the force exerted by the seat of the car on the passenger is 790 N.

Question 9

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 sum of the forces exerted on the ball?
A)0 N
B)20 N
C)40 N
D)50 N
E)100 N

Accepted Answer

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

Question 10

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 force that the string exerts on the bead if the angle α is measured to be 25°? 
A)0.089 N
B)0.041 N
C)0.20 N
D)0.22 N
E)0.46 N

Accepted Answer

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

Question 11

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?
A)9000 N
B)9800 N
C)300 N
D)900 N
E)3000 N

Accepted Answer

The friction force required to keep the car from sliding can be found 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 0.20 km, so the radius is 0.10 km or 100 meters. Plugging in the values, $F = \frac{1000 \times 30^2}{100} = 9000 N$.

Question 12

In order to simulate apparent weightlessness for astronauts in training (when the astronauts do not push on a seat),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?
A)79 m/s
B)310 m/s
C)260 m/s
D)160 m/s
E)510 m/s

Accepted Answer

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

Question 13

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 force that the string can exert on an object without breaking is 350 N,what is the maximum speed the toy can have without breaking the string?
A)700 m/s
B)26 m/s
C)19 m/s
D)13 m/s

Accepted Answer

The maximum tension force that the string can exert on the toy is 350 N. The centripetal force required to keep the toy moving in a circle is given by:$$F_c = mv^2/r$$where m is the mass of the toy, v is its speed, and r is the radius of the circle.Setting the centripetal force equal to the maximum tension force, we get:$$mv^2/r = 350 N$$Solving for v, we get:$$v = \sqrt{350 N * r/m}$$Substituting the given values, we get:$$v = \sqrt{350 N * 1.0 m / 0.50 kg} = 26 m/s$$

Question 14

A car traveling at a steady 20 m/s rounds an 80-m radius horizontal unbanked curve with the tires on the verge of slipping.What is the maximum speed with which this car can round a second unbanked curve of radius 320 m if the coefficient of static friction between the car's tires and the road surface is the same in both cases?
A)160 m/s
B)80 m/s
C)70 m/s
D)40 m/s
E)30 m/s

Accepted Answer

The maximum speed with which the car can round the first curve can be found using the formula $v=\sqrt{\mu g r}$, where $\mu$ is the coefficient of static friction, $g$ is acceleration due to gravity, and $r$ is the radius of the curve. Thus, $v = \sqrt{\mu g r} = \sqrt{(0.5)(9.8 m/s^2)(80m)} \approx 20m/s$. Since the car is already at its maximum speed for the first curve, it must slow down in order to stay within the limits of friction. The maximum speed for the second curve can be found using the same formula, but with the radius of the second curve: $v=\sqrt{\mu g r} = \sqrt{(0.5)(9.8 m/s^2)(320m)} \approx 40m/s$. Therefore, the answer is D.

Question 15

A future use of space stations may be to provide hospitals for severely burned persons.It is very painful for a badly burned person on Earth to lie in bed.In a space station,the effect of gravity can be reduced or even eliminated.How long should each rotation take for a doughnut-shaped hospital of 200-m radius so that persons on the outer perimeter would experience 1/10 the normal gravitational pull of Earth?
A)91 min
B)8.7 min
C)4.6 min
D)1.5 min
E)0.011 min

Accepted Answer

The answer of A future use of space stations may...

Question 16

A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diameter 1.0 m.The ball makes 2.0 revolutions every 1.0 s.
(a)What are the magnitude and direction of the acceleration of the ball?
(b)Find the tension force that the string exerts on the ball.

Accepted Answer

The answer of A small 175-g ball on the end...

Question 17

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 (the force that the pilot exerts on the seat)of the 70-kg pilot at that point?
A)3300 N
B)690 N
C)2600 N
D)490 N
E)1400 N

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, plus the force of gravity. The centripetal force (Fc) is given by $F_c = \frac{mv^2}{r}$, where $m$ is the mass, $v$ is the velocity, and $r$ is the radius of the circle. The force of gravity (Fg) acting on the pilot is $F_g = mg$, where $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$). The apparent weight is the sum of these two forces.Given:- $m = 70 \, kg$- $v = 150 \, m/s$- $r = 600 \, m$- $g = 9.8 \, m/s^2$First, calculate the centripetal force:$$F_c = \frac{mv^2}{r} = \frac{70 \times 150^2}{600} = \frac{70 \times 22500}{600} = \frac{1575000}{600} = 2625 \, N$$Then, calculate the force of gravity:$$F_g = mg = 70 \times 9.8 = 686 \, N$$The apparent weight is the sum of $F_c$ and $F_g$:$$2625 \, N + 686 \, N = 3311 \, N$$Rounding to the nearest hundred gives approximately $3300 \, N$, which matches choice A.

Question 18

A 20-g bead is attached to a light 120 cm-long string as shown in the figure.If the angle α is measured to be 18°,what is the speed of the bead? 
A)0.55 m/s
B)2.0 m/s
C)3.8 m/s
D)1.3 m/s
E)1.1 m/s

Accepted Answer

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

Question 19

A jet plane flying 600 m/s has 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?
A)640 m
B)1200 m
C)7100 m
D)9200 m

Accepted Answer

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

Question 20

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?
A)8.9 m/s
B)2.3 m/s
C)3.0 m/s
D)4.9 m/s
E)It depends on the mass of the passengers.

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 static friction, which is $f_s = \mu_s N$, where $\mu_s$ 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 static friction equal to the gravitational force (since it must counteract gravity to prevent the passengers from falling), we get $\mu_s \frac{mv^2}{r} = mg$, where $g$ is the acceleration due to gravity (9.8 m/s$^2$). Solving for $v$, we find $v = \sqrt{\frac{gr}{\mu_s}}$. Substituting $g = 9.8$ m/s$^2$, $r = 2.0$ m, and $\mu_s = 0.25$, 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.

Quiz 5: Circular Motion

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?

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?

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 force exerted by the seat of the car on a 60.0-kg passenger when the car is at the top of the bump?

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 sum of the forces exerted on the motorcycle?

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 sum of the forces exerted on the ball?

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 force that the string exerts on the bead if the angle α is measured to be 25°?

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?

In order to simulate apparent weightlessness for astronauts in training (when the astronauts do not push on a seat) ,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?

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 (the force that the pilot exerts on the seat) of the 70-kg pilot at that point?

A 20-g bead is attached to a light 120 cm-long string as shown in the figure.If the angle α is measured to be 18°,what is the speed of the bead?

A jet plane flying 600 m/s has 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?