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Correct option-1 Concept: INSTANTANEOUS VELOCITY
\(v=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta x}{\Delta t}\)
Circular Motion
Circular motion is classified into two categories- 1.Uniform circular motion 2.Non-uniform circular motion Explanation: If a particle moves with a constant speed in a circle, then the motion is called uniform circular motion. In uniform-circular motion, particle possesses only centripetal acceleration (ar).
In a circular motion, the angular vector is always perpendicular to centripetal acceleration because:-
Hence, option-1 is correct. India’s #1 Learning Platform Start Complete Exam Preparation
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By the end of this section, you will be able to do the following:
The learning objectives in this section will help your students master the following standards:
What exactly do we mean by circular motion or rotation? Rotational motion is the circular motion of an object about an axis of rotation. We will discuss specifically circular motion and spin. Circular motion is when an object moves in a circular path. Examples of circular motion include a race car speeding around a circular curve, a toy attached to a string swinging in a circle around your head, or the circular loop-the-loop on a roller coaster. Spin is rotation about an axis that goes through the center of mass of the object, such as Earth rotating on its axis, a wheel turning on its axle, the spin of a tornado on its path of destruction, or a figure skater spinning during a performance at the Olympics. Sometimes, objects will be spinning while in circular motion, like the Earth spinning on its axis while revolving around the Sun, but we will focus on these two motions separately.
[BL][OL] Explain the difference between circular and rotational motions by using the Earth’s rotation about its axis and its revolution about the Sun. Explain that Earth’s rotation is slightly elliptical, although it is very nearly circular. [OL][AL] Ask students to come up with examples of circular motion. When solving problems involving rotational motion, we use variables that are similar to linear variables (distance, velocity, acceleration, and force) but take into account the curvature or rotation of the motion. Here, we define the angle of rotation, which is the angular equivalence of distance; and angular velocity, which is the angular equivalence of linear velocity. When objects rotate about some axis—for example, when the CD in Figure 6.2 rotates about its center—each point in the object follows a circular path.
The arc length, , is the distance traveled along a circular path. The radius of curvature, r, is the radius of the circular path. Both are shown in Figure 6.3.
Consider a line from the center of the CD to its edge. In a given time, each pit (used to record information) on this line moves through the same angle. The angle of rotation is the amount of rotation and is the angular analog of distance. The angle of rotation Δθ Δθ is the arc length divided by the radius of curvature. Δθ= Δs r Δθ= Δs r The angle of rotation is often measured by using a unit called the radian. (Radians are actually dimensionless, because a radian is defined as the ratio of two distances, radius and arc length.) A revolution is one complete rotation, where every point on the circle returns to its original position. One revolution covers 2π 2π radians (or 360 degrees), and therefore has an angle of rotation of 2π 2π radians, and an arc length that is the same as the circumference of the circle. We can convert between radians, revolutions, and degrees using the relationship 1 revolution = 2π 2π rad = 360°. See Table 6.1 for the conversion of degrees to radians for some common angles. 2π rad = 360° 1rad = 360° 2π ≈57.3° 2π rad = 360° 1rad = 360° 2π ≈57.3° 6.1
[BL] Review displacement, speed, velocity, acceleration. [AL] Ask students whether or not velocity changes in uniform circular motion. What about speed? What about acceleration? How fast is an object rotating? We can answer this question by using the concept of angular velocity. Consider first the angular speed (ω) (ω) is the rate at which the angle of rotation changes. In equation form, the angular speed is ω= Δθ Δt , ω= Δθ Δt , 6.2 which means that an angular rotation (Δθ) (Δθ) occurs in a time, Δt Δt . If an object rotates through a greater angle of rotation in a given time, it has a greater angular speed. The units for angular speed are radians per second (rad/s). Now let’s consider the direction of the angular speed, which means we now must call it the angular velocity. The direction of the angular velocity is along the axis of rotation. For an object rotating clockwise, the angular velocity points away from you along the axis of rotation. For an object rotating counterclockwise, the angular velocity points toward you along the axis of rotation. Angular velocity (ω) is the angular version of linear velocity v. Tangential velocity is the instantaneous linear velocity of an object in rotational motion. To get the precise relationship between angular velocity and tangential velocity, consider again a pit on the rotating CD. This pit moves through an arc length (Δs) (Δs) in a short time (Δt) (Δt) so its tangential speed is v= Δs Δt . v= Δs Δt . 6.3 From the definition of the angle of rotation, Δθ= Δs r Δθ= Δs r , we see that Δs=rΔθ Δs=rΔθ . Substituting this into the expression for v gives v= rΔθ Δt =rω . v= rΔθ Δt =rω . The equation v=rω v=rω says that the tangential speed v is proportional to the distance r from the center of rotation. Consequently, tangential speed is greater for a point on the outer edge of the CD (with larger r) than for a point closer to the center of the CD (with smaller r). This makes sense because a point farther out from the center has to cover a longer arc length in the same amount of time as a point closer to the center. Note that both points will still have the same angular speed, regardless of their distance from the center of rotation. See Figure 6.4.
[AL] Explain that the time period Δt Δt in the equation that defines tangential velocity ( v= Δs Δt v= Δs Δt ) must be short so that the arc described by the moving object can be approximated as a straight line. This allows us to define the direction of the tangential velocity as being tangent to the circle. This approximation becomes increasingly accurate as Δt Δt becomes increasingly small. Now, consider another example: the tire of a moving car (see Figure 6.5). The faster the tire spins, the faster the car moves—large ω ω means large v because v=rω v=rω . Similarly, a larger-radius tire rotating at the same angular velocity, ω ω, will produce a greater linear (tangential) velocity, v, for the car. This is because a larger radius means a longer arc length must contact the road, so the car must move farther in the same amount of time.
However, there are cases where linear velocity and tangential velocity are not equivalent, such as a car spinning its tires on ice. In this case, the linear velocity will be less than the tangential velocity. Due to the lack of friction under the tires of a car on ice, the arc length through which the tire treads move is greater than the linear distance through which the car moves. It’s similar to running on a treadmill or pedaling a stationary bike; you are literally going nowhere fast.
Angular velocity ω and tangential velocity v are vectors, so we must include magnitude and direction. The direction of the angular velocity is along the axis of rotation, and points away from you for an object rotating clockwise, and toward you for an object rotating counterclockwise. In mathematics this is described by the right-hand rule. Tangential velocity is usually described as up, down, left, right, north, south, east, or west, as shown in Figure 6.6.
This video reviews the definition and units of angular velocity and relates it to linear speed. It also shows how to convert between revolutions and radians.
Click to view content For an object traveling in a circular path at a constant angular speed, would the linear speed of the object change if the radius of the path increases?
In this activity, you will create and measure uniform circular motion and then contrast it with circular motions with different radii.
If you swing an object slowly, it may rotate at less than one revolution per second. What would be the revolutions per second for an object that makes one revolution in five seconds? What would be its angular speed in radians per second?
Now that we have an understanding of the concepts of angle of rotation and angular velocity, we’ll apply them to the real-world situations of a clock tower and a spinning tire.
The clock on a clock tower has a radius of 1.0 m. (a) What angle of rotation does the hour hand of the clock travel through when it moves from 12 p.m. to 3 p.m.? (b) What’s the arc length along the outermost edge of the clock between the hour hand at these two times?
We can figure out the angle of rotation by multiplying a full revolution ( 2π
2π radians) by the fraction of the 12 hours covered by the hour hand in going from 12 to 3. Once we have the angle of rotation, we can solve for the arc length by rearranging the equation Δθ= Δs r
Δθ= Δs r since the radius is given.
In going from 12 to 3, the hour hand covers 1/4 of the 12 hours needed to make a complete revolution. Therefore, the angle between the hour hand at 12 and at 3 is 1 4 ×2πrad= π 2 1 4 ×2πrad= π 2 (i.e., 90 degrees).
Rearranging the equation Δθ= Δs r , Δθ= Δs r , we get Inserting the known values gives an arc length of Δs = ( 1.0m )( π 2 rad ) = 1.6m Δs = ( 1.0m )( π 2 rad ) = 1.6m
We were able to drop the radians from the final solution to part (b) because radians are actually dimensionless. This is because the radian is defined as the ratio of two distances (radius and arc length). Thus, the formula gives an answer in units of meters, as expected for an arc length.
Calculate the angular speed of a 0.300 m radius car tire when the car travels at 15.0 m/s (about 54 km/h). See Figure 6.5.
In this case, the speed of the tire tread with respect to the tire axle is the same as the speed of the car with respect to the road, so we have v = 15.0 m/s. The radius of the tire is r = 0.300 m. Since we know v and r, we can rearrange the equation v=rω
v=rω , to get ω= v r
ω= v r and find the angular speed.
To find the angular speed, we use the relationship: ω= v r ω= v r . Inserting the known quantities gives ω = 15.0m/s 0.300m = 50.0rad/s. ω = 15.0m/s 0.300m = 50.0rad/s.
When we cancel units in the above calculation, we get 50.0/s (i.e., 50.0 per second, which is usually written as 50.0 s−1). But the angular speed must have units of rad/s. Because radians are dimensionless, we can insert them into the answer for the angular speed because we know that the motion is circular. Also note that, if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular speed of ω = 15.0m/s 1.20m = 12.5rad/s ω = 15.0m/s 1.20m = 12.5rad/s 1.
What is the angle in degrees between the hour hand and the minute hand of a clock showing 9:00 a.m.? 2.
What is the approximate value of the arc length between the hour hand and the minute hand of a clock showing 10:00 a.m if the radius of the clock is 0.2 m?
3.
What is circular motion?
4.
What is meant by radius of curvature when describing rotational motion?
5.
What is angular velocity?
6.
What equation defines angular velocity, ω when r is the radius of curvature, θ is the angle, and t is the time?
7.
Identify three examples of an object in circular motion.
8.
What is the relative orientation of the radius and tangential velocity vectors of an object in uniform circular motion?
Use the Check Your Understanding questions to assess whether students master the learning objectives of this section. If students are struggling with a specific objective, the formative assessment will help identify which objective is causing the problem and direct students to the relevant content. |