Finding Angular Speed of a Disc?
Apply conservation of angular momentum principle. The angular momentum is additive, that means the angular momentum of the whole disk-box system equals angular momentum of the disk plus angular momentum of the box w.r.t. to the axis of rotation. L = L_disk + L_box Since the angular momentum equals monet of inertia times angular speed and both object move with same angular velocity around the axis of rotation you get L = (J_disk + J_box)∙ω The moment of inertia for an disk of radius R and mass m₁ rotating in the described way is J_disk = (1/2)∙m₁∙R² The angular momentum momentum of a small object of mass m₂ rotating at distance r to the axis of rotation is J_box = m₂∙r³ So the angular momentum of disk and box, at the instant, when box is at distance r to the axis is: L = ((1/2)∙m₁∙R² + m₂∙r³)∙ω Initially the box is at the center of the disk, i.e. at at r=0 and the disk rotates with speed ω₀. So the initial angular momentum is L₀ = ((1/2)∙m₁∙R² + m₂∙0³)∙ω₀ = (1/2)∙m₁∙R²∙ω₀ After the box has stopped it has reached r=R/2 and the disk rotates with unknown angular speed ω₁. Then the angular momentum is: L₁ = ((1/2)∙m₁∙R² + m₂∙(R/2)³)∙ω₁ = ((1/2)∙m₁∙R² + (1/4)∙m��∙R²)∙ω₁ = ((1/2)∙m₁ + (1/4)∙m₂)∙R²∙ω₁ Conservation of angular momentum requires that: L₀ = L₁ <=> (1/2)∙m₁∙R²∙ω₀ = ((1/2)∙m₁ + (1/4)∙m₂)∙R²∙ω₁ Hence at the end of box' motion out of the center of the disk the disk rotates with angular speed ω₁ = ω₀∙ (1/2)∙m₁∙R²∙/ ( ((1/2)∙m₁ + (1/4)∙m₂)∙R² )∙ = ω₀∙ m₁/(m₁ + (1/2)∙m₂)
Center of mass/speed/angular speed?
a) because PE=KElinear+KErotational, you can do some algebra to mgy=1/2mv^2+1/2IW^2. becomes mgy=1/2mv^2+1/1(1/2mr^2)(V/r)^2... and solve for v, you'll get v=square root of 4/3gy. your y (opposite over hypotenuse) will be sin angle (15.5)=y/ramp length(4.3). so y=4.3 sin 15.5. b) find the Circumference of the disk. C=2 pi r. that is one revolution down the ramp. but we need to compare that with speed, which you found for part a. V/C= answer in rev/sec. multiply that by 2pi to get it into rad/sec. (assuming you're answer format was the same as mine, on webassign... you're probably in my class :) )good luck!
Angular Speed of Disc?
Approach: Use the principle of Conservation of Angular Momentum: (1) Li = Lf, where: (2) L = angular momentum = I * ω, where: (3) I = moment of inertia The initial Li is given by: (4) Li = Ii * ω0, where: (5) Ii = the initial moment of inertia (MoI) of the disk about the z axis = Idiskz = m1 * R² / 2 [see Source 1] (We'll ignore the small block of mass m2 because it's at the center, making its radius = 0.) Substituting (5) into (4) we get: (6) Li = (m1 * R² / 2) * ω0 The final MoI is equal to Idiskz above plus the MoI of the block at distance R/2, because MoI's are additive: (7) If = Idiskz + m2 * (R / 2)² [because MoI of point-mass at radius r = m * r²] = (m1 * R² / 2) + m2 * R² / 4 = (R² / 2) * (m1 + m2 / 2) Using (1), (6) and (7): (8) (m1 * R² / 2) * ω0 = (R² / 2) * (m1 + m2 / 2) * ωf ► (9) m1 * ω0 = (m1 + m2 / 2) * ωf ► (10) ωf = m1 * ω0 / (m1 + m2 / 2) <<<=== Angular speed when block stops at R/2 .
Angular speed?
For a) The moment of inertia of a disk is I = 1/2 M r^2, so for the disk it would be I = 102/2 * 2.1^2 = 225 kg m^2 total angular momentum is zero e.g. omega_disk * I_disk = omega_person * I_person I_person = 50 kg * (1.2 m)^2 = 72 kg m^2 so, omega_disk * 225 kg m^2 = 72 kg m^2 * omega_person Furthermore, the velocity relative to the ground of the person is 2.1 m/s, so since v = omega *r omega_person * 1.2 m = 2.1 m/s -> omega_person = 1.75 rad/sec Thus, we have omega_disk = 1.75 *72 / 225 = 0.56 rad/sec spinning in the opposite direction from the direction of the person For part b), we use the sum of the two angular frequencies to find the time omega_disk+omega_person = 0.56 + 1.75 = 2.31 rad/sec Therefore, to go 2pi radians would take 2.72 seconds
Angular Speed on a merry go round ?
A merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.4 meters, and a mass M = 271 kg. A small boy of mass m = 47 kg runs tangentially to the merry-go-round at a speed of v = 2.1 m/s, and jumps on. R = 1.4 meters M = 271 kg m = 47 kg v = 2.1 m/s Q1: Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round. Q2:Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy. Q3.The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry-go-round when the boy is half way between the edge and the center of the merry go round? Q4.The boy then crawls to the center of the merry-go-round. What is the angular speed in radians/second of the merry-go-round when the boy is at the center of the merry go round? Q5. Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow, he manages to jump in such a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed in radians/second of the merry-go-round after the boy jumps off? So initially I did mvr = Iw where I was = to .5(M+m)R^2 but this wasn't right I'm lost. Thank you for your assistance, time and explanations
How to find angular speed, linear speed, centripetal acceleration and total distance?
Nope. The angular velocity (velocity) is continuous over all radii of the MGR. call it w. So the gal on the sting is spinning on the comparable cost because of the fact the guy close to the middle...w. Then the linear, tangential velocity (velocity) is v = wr close to the middle of the MGR and V = wR close to the sting, the place R > r. So the guy on the sting has a quicker tangential velocity, however the comparable angular velocity. Like linear velocity, angular velocity is made up of importance (radians according to 2d) and path (CCW and CW). Spin, rotation, angular velocity, action around the axis, and so on. all advise exceedingly plenty the comparable subject.
Physics: what is the angular and linear speed..?
enable s = the dimensions of a around arc formed with the help of an perspective theta measured in radians. enable r = radius of circle. Then s = r*theta. Dividing the two components with the help of t (time) supplies s/t = r*theta/t. s/t is the linear velocity and theta/t is the angular velocity. So v = r* theta/t relates linear and angular velocity. one million revolution = 2pi radians So distance traveled consistent with revolution is 2pi* 3.5 = 7pi ft. In 2 minutes the wheel makes 20 revolutions. for that reason distance traveled = 20*7pi = one hundred forty pi ft = approx 439.8 ft. The linear velocity is 439.8/a hundred and twenty = 3.sixty seven ft/2d. The angular velocity is 2pi/6 = pi/3 radians/sec
The angular momentum for a massive object with velocity v is maximal when the velocity is orthogonal to the radial arm and zero when the velocity v is aligined with the radial arm. Only one mathematical thingy fits this and it is the cross product of the radial arm with the velocity arm multiplied by the mass. And that is why the angular momentum is orthogonal to both the radial arm and the linear momentum vector. If you want to loosen up a stubborn nut push tangentially on your wrench, not radially.
A disk rotates with constant angular acceleration. The initial angular speed of the disk is 2π rad/s. After th?
A disk rotates with constant angular acceleration. The initial angular speed of the disk is 2π rad/s. After th? A disk rotates with constant angular acceleration. The initial angular speed of the disk is 2π rad/s. After the disk rotates through 18π radians, the angular speed is 9π rad/s. (a) What is the magnitude of the angular acceleration? (b) How much time did it take for the disk to rotate through 18π radians? (c) What is the tangential acceleration of a point located at a distance of 4.4 cm from the center of the disk?
A 51.4-cm diameter disk rotates with a constant angular acceleration...?
As the disk rotates, its angular velocity increases at the rate of 2.5 rad/s each second. To determine its angular velocity at 2.3 seconds, use the following equation. ωf = ωi + α * t, ωi = 0 ωf = 2.5 * 2.3 = 5.75 rad/s Since point P is on the rim of the disk, the distance from the center to this point is the radius of the disk. r = 0.5 * 0.514 = 0.257 meter To convert angular velocity to linear velocity, multiply by this number. v = 5.75 * 0.257 = 1.47775 m/s Its linear velocity increased from 0 m/s to 1.47775 in 2.3 seconds. To determine its tangential acceleration, divide its linear velocity by the time. a = 1.47775 ÷ 2.3 = 0.62425 m/s^2 Use the following equation to determine the angle the point has rotated in 2.3 seconds. θ = ½ * (ωi + ωf) * t, ωi = 0 θ = ½ * 5.75 * 2.3 = 6.6125 radians One revolution is 2 π radians. One radian = 360˚ ÷ (2 * π) = 180˚/π This is approximately 57.3˚. To determine the angle the point has rotated in 2.3 seconds, use the following equation. θ = 6.6125 radians * 180˚/π degrees/ radian θ = 1190.25/π degrees This is approximately 378.9˚ To determine the angle respect to the positive x-axis, subtract 360˚. θ = 1190.25/π – 360 This is approximately 18.9˚. The total angle is 57.3˚ plus this number. OR Since 57.3˚ is one radian, the total angle is 7.6125 radians 7.6125 * 180˚/π = 1370.25/π This is approximately 436.2˚ 1370.25/π – 360 This is approximately 76.2. This is same answer.