Physics 8 MC

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.260 rev/s . The magnitude of the angular acceleration is 0.893 rev/s2 . Both the the angular velocity and angular accleration are directed clockwise. The electric ceiling fan blades form a circle of diameter 0.710 m . What is the magnitude a of the resultant acceleration of a point on the tip of the blade at time t = 0.193 s ?

Centripetal acceleration acent for an object moving with tangential velocity vtan in a circular path of radius r is given by acent=v2tan/r.

radial acceleration

Ar=rw²

A merry-go-round is rotating at constant angular speed. Two children are riding the merry-go-round: Ana is riding at point A and Bobby is riding at point B. point b is closer to the outside, point a is closer to the axis. rotates at constant speed. Who moves with greater magnitude of tangential acceleration? Who moves with greater magnitude of tangential acceleration? Ana has the greater magnitude of tangential acceleration. Bobby has the greater magnitude of tangential acceleration. Both Ana and Bobby have the same magnitude of tangential acceleration.

Both Ana and Bobby have the same magnitude of tangential acceleration. Both Ana and Bobby are maintaining a constant speed, so they both have a tangential acceleration of zero (thus they are equal)! at=rα α=∆w/∆t

Who has the greater magnitude of centripetal acceleration? Who has the greater magnitude of centripetal acceleration? Ana has the greater magnitude of centripetal acceleration. Bobby has the greater magnitude of centripetal acceleration. Both Ana and Bobby have the same magnitude of centripetal acceleration.

... Bobby has the greater magnitude of centripetal acceleration

An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in time t, through what angle did it rotate in the time 12t? If this object rotates through an angle in time , through what angle did it rotate in the time ? θ. 4θ. 1/2θ. 1/4θ. 2θ.

1/4θ. θ=wi +1/2αt² wi=0 (1/2t)² = 1/4t²×1/2 = 1/8, then cancel out 1/2 from original equation where t is t

tangental acceleration

At=rα

The cans have essentially the same size, shape, and mass. Which can has more energy at the bottom of the ramp? Ignore friction and air resistance. The cans have essentially the same size, shape, and mass. Which can has more energy at the bottom of the ramp? Ignore friction and air resistance. The milk The refried beans They both have the same energy. The milk and the refried beans start out with the same amount of gravitational potential energy. Since mechanical energy is conserved in this experiment, both the milk and the refried beans must have the same amount of energy at the bottom of the ramp as well, but it may be divided differently between rotational kinetic energy and translational kinetic energy.

Correct The milk and the refried beans start out with the same amount of gravitational potential energy. Since mechanical energy is conserved in this experiment, both the milk and the refried beans must have the same amount of energy at the bottom of the ramp as well, but it may be divided differently between rotational kinetic energy and translational kinetic energy. They both have the same energy.

Consider a force F=80N applied to a beam as shown in the figure below. The length of the beam is ℓ=5.0m, and θ=37∘, so that x=3.0m and y=4.0m. Of the following expressions, which ones give the correct torque produced by the force F→ around point P? Of the following expressions, which ones give the correct torque produced by the force around point ? (80 N)(5.0 m). (48 N)(5.0 m)(sin 37∘). 80 N . (80 N)(5.0 m)(sin 37∘)). (48 N)(5.0 m). (80 N)(4.0 m). (80 N)(3.0 m).

Equations τ = rFsinθ show that there are three ways in which the torque can be written. It can be the product of the force, the lever arm, and the sine of the angle between them as in answer (c). It can be the product of the force and the component of the lever arm perpendicular to the force, as in answer (e). It can also be written as the product of the lever arm and the force perpendicular to the lever arm, as in answer (f). Doing the calculations shows that all three torques are equal. (80 N)(5.0 m)(sin 37∘)). (48 N)(5.0 m). (80 N)(3.0 m).

The experimenter from the video rotates on his stool, this time holding his empty hands in his lap. You stand on a desk above him and drop a long, heavy bean bag straight down so that it lands across his lap, in his hands. What happens?

He spins slower. Correct The bean bag is not rotating before it contacts the experimenter, so (by conservation of rotational momentum) he must supply the rotational momentum required to make it spin at the same rate as his lap. Therefore, he ends up spinning slower.

Two spheres have the same radius and equal mass. One sphere is solid, and the other is hollow and made of a denser material. Which one has the bigger moment of inertia about an axis through its center? The solid one. The hollow one. Both the same.

I=1/2McmR² where mass is located is important. The hollow one. The location of the mass is very important. Imagine taking the material from the solid sphere and compressing it outward to turn the solid sphere into a hollow sphere of the same mass and radius. As you do this you would be moving mass farther away from the axis of rotation, which will increase the moment of inertia. Therefore the hollow sphere has a greater moment of inertia than the solid sphere.

A small mass m on a string is rotating without friction in a circle. The string is shortened by pulling it through the axis of rotation without any external torque, see the figure What happens to the tangential velocity of the object? What happens to the tangential velocity of the object? It increases. It decreases. It remains the same.

It increases Work is done on the object, and so its kinetic energy increased. Thus the tangential velocity had to increase. Another way to consider the problem is that KE = L2/2I−−−−−√. Thus the kinetic energy (and so the speed) had to increase Translational kinetic energy= 1/2Mvcm², so if KE increases, so must V

A small mass m on a string is rotating without friction in a circle. The string is shortened by pulling it through the axis of rotation without any external torque (Figure 1) . What happens to the angular velocity of the object? What happens to the angular velocity of the object? It increases. It decreases. It remains the same.

It increases. Without an external torque students may think that the angular speed would remain constant. But with no external torque, the angular momentum must remain constant. The angular momentum is the product of the moment of inertia and the angular speed. As the string is shortened, the moment of inertia of the block decreases. Thus, the angular speed increases. Angular momentum = L = Iw I=xMR², so depends on radius, therefore if radius decreases the w must increase.

If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance) If you used 1000 of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance) so that it wasn't rotating? so that it was also rotating? It makes no difference.

It takes energy to rotate the ball. If some of the 1000 J goes into rotation, less is available for linear kinetic energy, and so the rotating ball will travel slower. so that it wasn't rotating

A turntable is rotating at 3313rpm. You then flip a switch, and the turntable speeds up, with constant angular acceleration, until it reaches 78rpm. Suppose you are asked to find the amount of time t, in seconds, it takes for the turntable to reach its final rotational speed. Which of the following equations could you use to directly solve for the numerical value of t? Suppose you are asked to find the amount of time , in seconds, it takes for the turntable to reach its final rotational speed. Which of the following equations could you use to directly solve for the numerical value of ? θ=θ0+ω0t+12αt2 ω=ω0+αt ω2=ω20+2α(θ−θ0) More information is needed before t can be found.

More information is needed before t can be found. Knowing ω and ω0 would not be enough information to allow you to solve for t; you would need additional information (namely, the value of the angular acceleration α) before you could find a numerical value for t.

moment of inertia related to force

Objects with larger moments of inertia I are harder to rotate. Since all of the objects are initially rotating at the same speed (since they are traveling at the same speed and have the same radius), large-I objects will have more energy than low-I objects of equal mass.

A small solid sphere and a small thin hoop are rolling along a horizontal surface with the same translational speed when they encounter a 20 ∘ rising slope. Part A If these two objects roll up the slope without slipping, which will rise farther up the slope? If these two objects roll up the slope without slipping, which will rise farther up the slope? The hoop. The sphere. Both the same. More information about the objects' mass and diameter is needed.

The hoop. If you do not take into account the energy of rotation, you would answer that the two objects would rise to the same height. Another common misconception is that the mass and/or diameter of the objects will affect how high they travel. When using conservation of energy to relate the total initial kinetic energy (translational and rotational) to the final potential energy, the mass and radius of the objects cancel out. The thin hoop has a larger moment of inertia (for a given mass and radius) than the solid sphere. It will therefore have a greater total initial kinetic energy, and will travel to a greater height on the ramp. MR² Vs. 2/5MR²

Which of the following statements are correct? children spinning in circle, last child runs furthest. Check all that apply. The last child in the line has the greatest tangential acceleration. The last child in the line has the greatest radial acceleration. All the children have the same tangential acceleration. All the children have the same radial acceleration.

The last child in the line has the greatest tangential acceleration. The last child in the line has the greatest radial acceleration. The last child in the line, being the farthest away from the axis of rotation, has the greatest radial acceleration. The force needed to produce this acceleration is provided by the pull of the rest of the children in the line. It won't take long before this acceleration becomes too high, especially for the final two children in the line. At this point, those children will not be able to apply enough force to hold on and will have to let go.

Two wheels having the same radius and mass rotate at the same angular velocity ((Figure 1) ). One wheel is made with spokes so nearly all the mass is at the rim. The other is a solid disk. How do their rotational kinetic energies compare? The solid wheel has about twice the KE. They are nearly the same. The wheel with spokes has higher KE, but not twice as high. The wheel with spokes has about twice the KE. The solid wheel has higher KE, but not twice as high.

The wheel with spokes has about twice the KE. If you don't consider how the location of the mass affects the moment of inertia, you may think that the two kinetic energies are nearly the same. However, a hollow cylinder has twice the moment of inertia as a solid cylinder of the same mass and radius. The kinetic energy is proportional to the moment of inertia, so at the same angular speed the wheel with the spokes will have nearly double the kinetic energy as the solid cylinder. It is only nearly double because some of the mass is in the spokes so the moment of inertia is not exactly double. I of thin band=MR², I of solid disk= 1/2MR² Rotational KE= 1/2IW²

The solid dot shown in the figure below is a pivot point. The board can rotate about the pivot.

Torque is the product of the lever arm and the component of the force perpendicular to the arm. Although the 1000 N force has the greatest magnitude, it acts at the pivot and so the lever arm is zero, and thus the torque is also zero. The 800 N force is parallel to the lever arm and also exerts no torque. Of the three 500 N forces, (c) is both perpendicular to the lever arm and farthest from the pivot.

moment of inertia

determines the torque needed for a desired angular acceleration about a rotational axis. It depends on the body's mass distribution and the axis chosen, with larger moments requiring more torque to change the body's rotation.

If there were a great migration of people toward the Earth's equator, the length of the day would If there were a great migration of people toward the Earth's equator, the length of the day would decrease because of conservation of energy. increase because of conservation of energy. decrease because of conservation of angular momentum. increase because of conservation of angular momentum. remain unaffected.

increase because of conservation of angular momentum. No net torque acts on the Earth so the angular momentum is conserved. As people move toward the equator their distance from the Earth's axis increases. This increases the moment of inertia of the Earth. For angular momentum to be conserved the angular speed must decrease and so it will take longer for the Earth to complete a full rotation. Angular momentum= Iw (i with people in middle = point mass= MR², with people on outside more like a sphere, so 2/5MR²

velocity of cans?

milk greater b/c takes less energy to rotate.

Suppose you are sitting on a rotating stool holding a 2- mass in each outstretched hand. If you suddenly drop the masses, your angular velocity will increase. decrease. stay the same.

stay the same. Students may mistakenly reason that since no net torque acts on you and your moment of inertia decreases as the masses are released, your angular speed should increase. This reasoning is erroneous because the angular momentum of the system of you and the masses is conserved. As the masses fall they carry angular momentum with them. If you consider you and the masses as two separate systems, each with angular momentum from their moments of inertia and angular speed, it is easy to see that by dropping the masses, no net external torque acts on you and your moment of inertia does not change, so your angular speed will not change. The angular momentum of the masses also does not change until they hit the ground and friction (external torque) stops their motion.

Assume that the motor has accelerated the wheel up to an angular velocity ω1 with angular acceleration α in time t1. At this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of −5α. Find t2, the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity). Express your answer in terms of some or all of the following: ω1, α, and t1.

t2 = ω1/5α

Bonnie sits on the outer rim of a merry-go-round, and Jill sits midway between the center and the rim. The merry-go-round makes one complete revolution every 2 seconds. Jill's linear velocity is: Bonnie sits on the outer rim of a merry-go-round, and Jill sits midway between the center and the rim. The merry-go-round makes one complete revolution every 2 seconds. Jill's linear velocity is: half of Bonnie's. the same as Bonnie's. twice Bonnie's. one-quarter of Bonnie's. four times Bonnie's.

v=rw w=2πrf half of Bonnie's.

A car speedometer that is supposed to read the linear speed of the car uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the car instead, how will that affect the speedometer reading? The speedometer If larger-diameter tires are mounted on the car instead, how will that affect the speedometer reading? The speedometer will read high. will still read the speed accurately. will read low.

v=rw θ=wi+1/2αt² r=2πf will read low. if r increased then v will increase, but angular velocity will stay constant A common error is to think that increasing the radius of the tires would increase the speed measured by the speedometer. This is actually backwards. Increasing the size of the tires will cause the car to travel faster than it would with smaller tires, when the wheels have the same angular speed. Therefore the speed of the car will be greater than the speed measured by the speedometer.

angular acceleration

α=∆w/∆t

Who moves with greater magnitude of angular acceleration? Ana has the greater magnitude of angular acceleration. Bobby has the greater magnitude of angular acceleration. Both Ana and Bobby have the same magnitude of angular acceleration.

α=∆w/∆t both same, because both zero

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration α. The flywheel is assumed to be at rest at time t=0 in Parts A and B of this problem. Find the angle θ1 through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity ω1. Express your answer in terms of some or all of the following: ω1, α, and t1.

θ1 = ω1^2/2α

You are now given an additional piece of information: It takes five complete revolutions for the turntable to speed up from 3313rpm to 45rpm. Which of the following equations could you use to directly solve for the numerical value of the angular acceleration α? You are now given an additional piece of information: It takes five complete revolutions for the turntable to speed up from to . Which of the following equations could you use to directly solve for the numerical value of the angular acceleration ? θ=θ0+ω0t+1/2αt2 ω=ω0+αt ω2=ω20+2α(θ−θ0) More information is needed before α can be found.

ω2=ω20+2α(θ−θ0) now know θ?