Question 1

In the figure,point P is at rest when it is on the x-axis.The linear speed of point P when it reaches the y-axis is closest to

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Question 2

When a 2.75-kg fan,having blades 18.5 cm long,is turned off,its angular speed decreases uniformly from 10.0 rad/s to 6.30 rad/s in 5.00 s.
(a)What is the magnitude of the angular acceleration of the fan?
(b)Through what angle (in degrees)does it turn while it is slowing down during the 5.00 s?
(c)If its angular acceleration does not change,how long after it is turned off does it take the fan to stop.

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Question 3

Two forces act on a 55-kg object.One force has magnitude 65 N directed 59° clockwise from the positive x-axis,and the other has a magnitude 35 N at 32° clockwise from the positive y-axis.What is the magnitude of this object's acceleration?

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Question 4

A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle.Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.

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Question 5

A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill.If it takes 3.60 s for it to make the first complete revolution,how long will it take to make the next complete revolution?

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Question 6

Future space stations will create an artificial gravity by rotating.Consider a cylindrical space station 390 m diameter rotating about its central axis.Astronauts walk on the inside surface of the space station.What rotation period will provide "normal" gravity?

Accepted Answer

To provide "normal" gravity, the centripetal acceleration at the surface must equal the acceleration due to gravity on Earth ($g = 9.8 \, \text{m/s}^2$). The formula for centripetal acceleration is $a = \frac{v^2}{r}$, where $v$ is the linear velocity and $r$ is the radius of the circle. Since $v = \frac{2\pi r}{T}$, where $T$ is the period, we can substitute $v$ into the acceleration formula to get $a = \frac{4\pi^2 r}{T^2}$. Setting $a = g$ and solving for $T$ gives $T = \sqrt{\frac{4\pi^2 r}{g}}$. The radius $r$ is half the diameter, so $r = 195 \, \text{m}$. Plugging in the values gives $T = \sqrt{\frac{4\pi^2 \times 195}{9.8}} \approx 28 \, \text{s}$.

Question 7

A new 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 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.

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Question 8

The angular acceleration of a wheel is given in rad/s² by 45t³ - 11t⁴, where t is in seconds.If the wheel starts from rest at t = 0.00 s,when is the next time the wheel is at rest?

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Question 9

A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point located 8.00 cm from the axis of rotation of the centrifuge moves with a speed of 150 m/s when the centrifuge is at full speed. (a)What is the angular acceleration (in rad/s²)of the centrifuge as it spins up? (b)How many revolutions does the centrifuge make as it goes from rest to its final angular speed?

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Question 10

A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m.The coefficient of static friction between the tires and the road is 0.350.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 of mass $m$ on a horizontal surface, $N = mg$, where $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$). 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 $v$ gives $v = \sqrt{\mu_s g r}$. Substituting $\mu_s = 0.350$, $g = 9.8 \, \text{m/s}^2$, and $r = 100 \, \text{m}$ gives $v = \sqrt{0.350 \times 9.8 \times 100} = \sqrt{343} \approx 18.5 \, \text{m/s}$.

Question 11

A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 13.5 rad/s.What is the magnitude of the angular acceleration of the wheel?

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Question 12

A turbine blade rotates with angular velocity ω(t)= 2.00 rad/s- 2.1.00 rad/s³t².What is the angular acceleration of the blade at t = 9.10 s?

Accepted Answer

Explanation:The angular acceleration is the derivative of the angular velocity with respect to time:$$\alpha(t)=\frac{d\omega(t)}{dt}=\frac{d}{dt}(2.00	ext{ rad/s}-2.100	ext{ rad/s}^3t^2)$$$$=-4.200	ext{ rad/s}^3t$$At t = 9.10 s, the angular acceleration is:$$\alpha(9.10	ext{ s})=-4.200	ext{ rad/s}^3(9.10	ext{ s})=-38.2	ext{ rad/s}^2$$

Question 13

In the figure,a weightlifter's barbell consists of two identical uniform spherical masses each with radius 0.17 m and mass of 50 kg.The weights are connected by a 0.96-m uniform steel rod with a mass of 12 kg.Find the moment of inertia of the barbell about the axis through the center (see figure).

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Question 14

The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t)= (2.00 rad/s²)t + (1.00 rad/s⁴)t³. (a)Through how many radians does the wheel turn during the first 2.00 s of its motion? (b)What is the angular acceleration (in rad/s²)of the wheel at the end of the first 2.00 s of its motion?

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Question 15

In the figure,point P is at rest when it is on the x-axis.The time t,when P returns to the original position on the x-axis,is closest to

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Question 16

A machinist turns the power on to a grinding wheel,which is at rest at time t = 0.00 s.The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off.The wheel decelerates uniformly at 1.5 rad/s² until the wheel stops.In this situation,the time interval of angular deceleration (slowing down)is closest to:

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Question 17

A Ferris wheel has diameter of 10 m and makes one revolution in 8.0 seconds.A person weighing 670 N is sitting on one of the benches attached at the rim of the wheel.What is the apparent weight (that is,the normal force exerted on her by the bench)of the person as she passes through the highest point of her motion?

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Question 18

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface.What is the moment of inertia of this sphere about an axis through its center?

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Question 19

A slender uniform rod 100.00 cm long is used as a meter stick.Two parallel axes that are perpendicular to the rod are considered.The first axis passes through the 50-cm mark and the second axis passes through the 30-cm mark.What is the ratio of the moment of inertia through the second axis to the moment of inertia through the first axis?

Accepted Answer

The moment of inertia of a slender rod about an axis through its center (the 50-cm mark) is given by $I_{center} = \frac{1}{12}ML^2$, where $M$ is the mass of the rod and $L$ is its length. For the axis through the 30-cm mark, we use the parallel axis theorem: $I_{30cm} = I_{center} + Md^2$, where $d$ is the distance from the center to the 30-cm mark (20 cm or 0.2 m). Thus, $I_{30cm} = \frac{1}{12}ML^2 + M(0.2)^2$. The ratio of $I_{30cm}$ to $I_{center}$ includes the term $\frac{1}{12}ML^2$ in both, which cancels out, leaving the ratio dependent on the additional $M(0.2)^2$ term for $I_{30cm}$. This results in a ratio greater than 1, and through calculation, it can be shown that the ratio is 1.5, considering the distribution of mass and the use of the parallel axis theorem.

Question 20

A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. When the power to the grinder is turned off,the grinding wheel slows with constant angular acceleration and takes 80.0 s to come to a rest. (a)What was the angular acceleration (in rad/s²)of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive? (b)How many revolutions did the wheel make during the time it was coming to rest?

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Quiz 11: Motion in a Circle

In the figure,point P is at rest when it is on the x-axis.The linear speed of point P when it reaches the y-axis is closest to

Two forces act on a 55-kg object.One force has magnitude 65 N directed 59° clockwise from the positive x-axis,and the other has a magnitude 35 N at 32° clockwise from the positive y-axis.What is the magnitude of this object's acceleration?

A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle.Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.

A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill.If it takes 3.60 s for it to make the first complete revolution,how long will it take to make the next complete revolution?

Future space stations will create an artificial gravity by rotating.Consider a cylindrical space station 390 m diameter rotating about its central axis.Astronauts walk on the inside surface of the space station.What rotation period will provide "normal" gravity?

The angular acceleration of a wheel is given in rad/s² by 45t³ - 11t⁴, where t is in seconds.If the wheel starts from rest at t = 0.00 s,when is the next time the wheel is at rest?

A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m.The coefficient of static friction between the tires and the road is 0.350.At what speed will the car begin to skid sideways?

A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 13.5 rad/s.What is the magnitude of the angular acceleration of the wheel?

A turbine blade rotates with angular velocity ω(t) = 2.00 rad/s- 2.1.00 rad/s³t².What is the angular acceleration of the blade at t = 9.10 s?

In the figure,a weightlifter's barbell consists of two identical uniform spherical masses each with radius 0.17 m and mass of 50 kg.The weights are connected by a 0.96-m uniform steel rod with a mass of 12 kg.Find the moment of inertia of the barbell about the axis through the center (see figure).

In the figure,point P is at rest when it is on the x-axis.The time t,when P returns to the original position on the x-axis,is closest to

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface.What is the moment of inertia of this sphere about an axis through its center?

Quiz 11: Motion in a Circle

In the figure,point P is at rest when it is on the x-axis.The linear speed of point P when it reaches the y-axis is closest to

Two forces act on a 55-kg object.One force has magnitude 65 N directed 59° clockwise from the positive x-axis,and the other has a magnitude 35 N at 32° clockwise from the positive y-axis.What is the magnitude of this object's acceleration?

A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle.Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.

A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill.If it takes 3.60 s for it to make the first complete revolution,how long will it take to make the next complete revolution?

Future space stations will create an artificial gravity by rotating.Consider a cylindrical space station 390 m diameter rotating about its central axis.Astronauts walk on the inside surface of the space station.What rotation period will provide "normal" gravity?

The angular acceleration of a wheel is given in rad/s2 by 45t3 - 11t4, where t is in seconds.If the wheel starts from rest at t = 0.00 s,when is the next time the wheel is at rest?

A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m.The coefficient of static friction between the tires and the road is 0.350.At what speed will the car begin to skid sideways?

A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 13.5 rad/s.What is the magnitude of the angular acceleration of the wheel?

A turbine blade rotates with angular velocity ω(t) = 2.00 rad/s- 2.1.00 rad/s3 t2.What is the angular acceleration of the blade at t = 9.10 s?

In the figure,a weightlifter's barbell consists of two identical uniform spherical masses each with radius 0.17 m and mass of 50 kg.The weights are connected by a 0.96-m uniform steel rod with a mass of 12 kg.Find the moment of inertia of the barbell about the axis through the center (see figure).

In the figure,point P is at rest when it is on the x-axis.The time t,when P returns to the original position on the x-axis,is closest to

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface.What is the moment of inertia of this sphere about an axis through its center?

The angular acceleration of a wheel is given in rad/s² by 45t³ - 11t⁴, where t is in seconds.If the wheel starts from rest at t = 0.00 s,when is the next time the wheel is at rest?

A turbine blade rotates with angular velocity ω(t) = 2.00 rad/s- 2.1.00 rad/s³t².What is the angular acceleration of the blade at t = 9.10 s?