Question 1

A 2000 kg car is traveling on a banked curved icy road.The road is banked at an angle of 12.0 degrees and has a radius of curvature of 500 m.The velocity of the car necessary to travel on the icy road without sliding is

Accepted Answer

The velocity required to travel on a banked curve without sliding is given by the formula v = sqrt(r * g * tan(theta)), where r is the radius, g is the acceleration due to gravity, and theta is the banking angle. Substituting the given values, v = sqrt(500 * 9.8 * tan(12 degrees)) = 32.3 m/s.

Question 2

A 0.500 kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long.The stone is moving around the path at a constant frequency of 2.20 rev/sec.At the moment the stone is overhead,the stone is released.The magnitude and direction of the velocity of the stone when it leaves the circular path is

Accepted Answer

The velocity of the stone when released is the tangential velocity, which is horizontal at the top of the circle. The tangential velocity is calculated using v = 2πr * frequency, where r = 0.75 m and frequency = 2.20 rev/sec. This gives v = 10.4 m/s.

Question 3

A 5,000 kg satellite is orbiting the Earth in a circular path.The height of the satellite above the surface of the Earth is 800 km.The velocity of the satellite is (Me = 5.98 $\times$ 10²⁴ kg, Re = 6.37 $\times$ 10⁶ m, G = 6.67 $\times$ 10^-11 N.m²/kg²)

Accepted Answer

The correct answer is A because the velocity of the satellite can be calculated using the formula v = sqrt(GM/r), where G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth to the satellite. Plugging in the given values, we get v = sqrt((6.67 * 10^-11 N.m^2/kg^2) * (5.98 * 10^24 kg) / (6.37 * 10^6 m + 800 km)) = 7,460 m/s.

Question 4

A 4.00 kg mass is moving in a circular path of radius 3.20 m with a constant tangential velocity of 6.20 m/sec.The centripetal force on the mass is

Accepted Answer

The centripetal force can be calculated using the formula:

F = (m * v^2) / r

Where m is the mass of the object, v is its tangential velocity, and r is the radius of the circular path.

Plugging in the given values, we get:

F = (4.00 kg * (6.20 m/s)^2) / 3.20 m 
F = 48.1 N

Therefore, the centripetal force on the mass is 48.1 N, which corresponds to choice B.

Question 5

A 2000 kg car is traveling on a banked curved icy road without sliding.The velocity of the car is 30 m/s and the road is banked at an angle of 20.0 degrees.The radius of curvature of the road is

Accepted Answer

To solve this problem, we need to use the following formula for the radius of curvature of a banked curve:

R = (v^2 / g) * tan(θ)

where:
- R = radius of curvature
- v = velocity of the car
- g = acceleration due to gravity (9.8 m/s^2)
- θ = angle of the banked road

Plugging in the given values, we get:

R = (30^2 / 9.8) * tan(20)
R ≈ 252 m

Therefore, the best choice is (C) 252 m.

Question 6

A conical pendulum is constructed with a string 2.5 m in length.The pendulum is set in horizontal circular path about the vertical axis.If the angular velocity of the conical pendulum is a constant 2.8 rad/s,then the angle the string makes with the vertical axis is

Accepted Answer

The answer of A conical pendulum is constructed with a...

Question 7

A 2000 kg car is traveling on a banked curved icy road without sliding.The velocity of the car is 32.0 m/s and the road is banked at an angle of 20.0 degrees.The radius of curvature of the road is

Accepted Answer

The radius of curvature can be calculated using the formula for a banked curve without friction: $ r = \frac{v^2}{g \cdot \tan(\theta)} $. Substituting the given values: $ v = 32.0 \, \text{m/s} $, $ g = 9.8 \, \text{m/s}^2 $, and $ \theta = 20.0^\circ $, we find $ r \approx 287 \, \text{m} $.

Question 8

Two moons orbit a planet in nearly circular orbits.Moon A has orbital radius r,and moon B has orbital radius 4r.Moon A takes 20 days to complete one orbit.How long does it take moon B to complete one orbit?

Accepted Answer

According to Kepler's Third Law, the square of the period of a moon's orbit is proportional to the cube of its average distance from the planet. Therefore, we can set up the following proportion:
(period of B)^2/(average distance of B)^3 = (period of A)^2/(average distance of A)^3
Letting x be the period of B, we can simplify the proportion to:
x^2/(4r)^3 = (20)^2/r^3
Solving for x, we get:
x = 160 days.

Question 9

A 2.60 kg mass is moving in a circular path with a constant angular velocity of 5.50 rad/sec and with a tangential velocity of 3.50 m/sec.The centripetal force on the mass is

Accepted Answer

The answer of A 2.60 kg mass is moving in...

Question 10

A 4.00 kg mass is moving in a circular path of radius 4.10 m with a constant angular velocity of 5.00 rad/sec.The centripetal force on the mass is

Accepted Answer

The answer of A 4.00 kg mass is moving in...

Question 11

A conical pendulum is constructed with a string 2.00 m in length.The pendulum is set in horizontal circular path the vertical axis.If the angle the string makes with the vertical axis is 45.0 degrees,then the angular velocity of the conical pendulum is

Accepted Answer

The answer of A conical pendulum is constructed with a...

Question 12

A 1.00 kg stone attached to a 1.00 m long string is traveling in a vertical circular orbit.What is the minimum tangential velocity at the top of the vertical circular orbit to keep the string from going slack?

Accepted Answer

The minimum tangential velocity at the top of the circle is found using the formula v = sqrt(g*r), where g is the acceleration due to gravity (9.81 m/s²) and r is the radius (1.00 m). This gives v = sqrt(9.81*1) = 3.13 m/s.

Question 13

A 4.00 kg mass is moving in a circular path with a constant angular velocity of 5.00 rad/sec and with a tangential velocity of 5.00 m/sec.The centripetal force on the mass is

Accepted Answer

The answer of A 4.00 kg mass is moving in...

Question 14

A 0.500 kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long.The stone is moving around the path at a constant frequency of 1.50 rev/sec.At the moment the stone is overhead,the stone is released.The velocity of the stone when it leaves the circular path is

Accepted Answer

We can use conservation of energy to solve this problem. Initially, the stone has kinetic energy and potential energy due to its height. At the moment it is released, all of the potential energy is converted to kinetic energy.

The potential energy at the top of the circle is given by:
PE = mgh
where m is the mass of the stone, g is the acceleration due to gravity, and h is the height of the stone above some reference point (usually taken to be the lowest point in the circle). In this case, h = 0 and the potential energy is zero.

The kinetic energy of the stone is given by:
KE = (1/2)mv^2
where v is the speed of the stone.

At the bottom of the circle, the speed of the stone is given by:
v = sqrt(2gh)
where h is now the height of the stone above the lowest point in the circle. In this case, h = 2R (where R is the radius of the circle, which is equal to 0.75 m), so we have:
v = sqrt(2g(2R))
v = sqrt(2(9.81 m/s^2)(1.5 m))
v = 7.07 m/s

Therefore, the velocity of the stone when it leaves the circular path is 7.07 m/s, which is answer choice B.

Question 15

A 2000 kg car is traveling on a banked curved icy road.The velocity of the car is 25.0 m/s and the road has a radius of curvature of 500 m.If the car is to travel on the icy road without sliding,then the angle of the banked road is

Accepted Answer

The answer of A 2000 kg car is traveling on...

Question 16

A 5,000 kg satellite is orbiting the Earth in a geostationary orbit.The height of the satellite above the surface of the Earth is (Me = 5.98 $\times$ 10²⁴ kg, Re = 6.37 $\times$ 10⁶ m, G = 6.67 $\times$ 10^-11 N.m²/kg²)

Accepted Answer

A geostationary orbit has a radius of approximately 42,164 km from the Earth's center. Subtracting Earth's radius (6,370 km) gives a height of about 35,794 km or 3.59 * 10^7 m above the Earth's surface.

Question 17

A 5,000 kg satellite is orbiting the Earth in a circular path.The height of the satellite above the surface of the Earth is 800 km.The angular velocity of the satellite as it orbits the Earth is (Me = 5.98 $\times$ 10²⁴ kg, Re = 6.37 $\times$ 10⁶ m, G = 6.67 $\times$ 10^-11 N.m²/kg²)

Accepted Answer

The angular velocity ω can be calculated using the formula ω = sqrt(G * Me / r^3), where r is the distance from the center of the Earth to the satellite. Here, r = Re + height = 6.37 * 10^6 m + 800 * 10^3 m = 7.17 * 10^6 m. Plugging in the values, ω = sqrt(6.67 * 10^-11 * 5.98 * 10^24 / (7.17 * 10^6)^3) ≈ 1.04 * 10^-3 rad/s.

Question 18

A 5,000 kg satellite is orbiting the Earth in a circular path.The height of the satellite above the surface of the Earth is 800 km.The time it takes for the satellite to travel around the Earth is (Me = 5.98 $\times$ 10²⁴ kg, Re = 6.37 $\times$ 10⁶ m, G = 6.67 $\times$ 10^-11 N.m²/kg²)

Accepted Answer

We can use the formula T = 2π√(r³/GM), where T is the period of the satellite (the time it takes to complete one orbit), r is the radius of the orbit (which is equal to the sum of the radius of the Earth and the height of the satellite), G is the gravitational constant, M is the mass of the Earth.

r = Re + h = 6.37 × 10^6 m + 8 × 10^5 m = 7.17 × 10^6 m

Plugging in the values, we get:

T = 2π√((7.17 × 10^6)^3 / (6.67 × 10^-11 × 5.98 × 10^24))
T = 1.68 hours

Therefore, the correct answer is D) 1.68 hours.

Question 19

An airplane is traveling at 250 m/s in level flight.If the airplane is to make a change in direction,it must travel is a horizontal curved path.To fly in the curved path,the pilot banks the airplane at an angle such that the lift has a horizontal component that provides the horizontal centripetal acceleration to move in a horizontal circular path.If the airplane is banked at an angle of 15.0 degrees,then the radius of curvature of the curved path of the airplane is

Accepted Answer

The radius of curvature can be calculated using the formula $ r = \frac{v^2}{g \cdot \tan(\theta)} $, where $ v = 250 \, \text{m/s} $, $ g = 9.81 \, \text{m/s}^2 $, and $ \theta = 15.0^\circ $. Plugging in these values gives approximately 23.8 km.

Question 20

An airplane is traveling at 150 m/s in level flight.If the airplane is to make a change in direction,it must travel in a horizontal curved path.To fly in the curved path,the pilot banks the airplane at an angle such that the lift has a horizontal component that provides the horizontal centripetal acceleration to move in a horizontal circular path.If the airplane is banked at an angle of 12.0 degrees,then the radius of curvature of the curved path of the airplane is

Accepted Answer

The correct answer is A.The horizontal component of the lift provides the centripetal force necessary to keep the airplane moving in a circle. The radius of curvature of the path is given by:r = v^2 / (g * tan(theta))where:v is the speed of the airplaneg is the acceleration due to gravitytheta is the banking anglePlugging in the given values, we get:r = (150 m/s)^2 / (9.81 m/s^2 * tan(12.0 degrees)) = 10.8 km

Quiz 5: Circular Motion