A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.5 rad/s in 2.92 s.?
(11.5 rad/s)/(2.92 s) = 3.938 rad/s^2, that's the magnitude of the angular acceleration. The average angular speed is 5.75 rad/s, so the angle is (5.75 rad/s)(2.92 s) = 16.8 rad.
5) A wheel starts from rest and rotates with constant angular acceleration...?
(part 1 of 2) A wheel starts from rest and rotates with constant angular acceleration to an angular speed of 8.6 rad/s in 1.35 s. Find the magnitude of the angular acceler- ation of the wheel. Answer in units of rad/s2. (part 2 of 2) Find the angle in radians through which it rotates in this time. Answer in units of rad.
Wheel starts at rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s^2?
on condition that : preliminary ANGULAR-speed = 0 very final angular-speed, w = 15 rad/sec In time, t = 3 sec enable, angular-acceleration = a rad/s^2, SO, 15 = at = 3a, ===> a = 15/3 = 5, OR, a = 5 rad/sec^2 >=========================< answer
A wheel starts from rest and rotates with constant angular to reach an angular speed of 15.0 rad/s in 3.0s. Fi?
Use the equation: ?² =?? ² + 2 ? (?- ?? ) the position ? = very last angular velocity ?? = initial angular velocity (0 rad/sec hence) ? = angular acceleration ?- ?? = angular displacement (4 ? b/c 2 finished revolutions) (6 rad/s)² = (0 rad/s)² + 2 ? (4 ?) ? = a million.40 3 rad/s²
A wheel, starting from rest, rotates with a constant angular acceleration of 2.50 rad/s^2.?
a) Given: Initial Angular Velocity = 0 Angular Acceleration = 2.50 rad/sec^2 During a certain 4.50 sec interval, its angular displacement is 100.0 radians Find: t = time elapsed before the start of the 4.50 sec interval Solution: Angular Displacement = (Initial Angular Velocity)t + (1/2)(Angular Acceleration)t^2 θ = ωi(t) + (1/2)αt^2 100.0 rad = ωi(4.50s) + (1/2)(2.50rad/s^2)(4.50s)^2 ωi = [100 - (1/2)(2.50)(4.50)^2]/4.50 ωi = 16.6 rad/sec Now, starting from rest, ωi = 0, and ωf = 16.6 rad/sec with α = 2.5 rad/s^2. Hence, α = (ωf - ωi)/t t = (16.6 - 0)/2.50 = 6.64 sec The wheel has been turning for 6.64 sec before the start of the 4.50 sec interval. ANSWER b) Angular velocity of the wheel at start of 4.50 sec interval is 16.6 rad/sec (see solution above) ANSWER Hope this helps. teddy boy
A wheel starts from rest and rotates with constant angular acceleration to an angular speed of 8.6 rad/s in 2.88 s. PLEASE HELP!!!! PHYSICS?
A wheel starts from rest and rotates with constant angular acceleration to an angular speed of 8.6 rad/s in 2.88 s. A.) Find the magnitude of the angular acceleration of the wheel. ώ = Δω/Δt = 8.6/2.88 = 3.0 rad/sec^2 (angular acceleration) Θ = 1/2*ω*t = 4.3*2.88 = 12.4 rad ...corresponding to 4 turns approx Θ = 1/2*ώ*t^2 = 1.5*2.88^2 = 12.4 rad..same result as above: physics works!
A wheel starts from rest and has an angular acceleration that is given by alpha(t) = (6 rad/s^4)t^2...?
You go about this in exactly the same way you'd do a linear acceleration A over time T, to find out how far S one goes. That is exactly like S = UT + 1/2 AT^2; where U is the initial speed. But instead of meters or feet, linear distance, we use angular distance radians. And we have O = wT + 1/2 aT^2; where w is the initial angular speed (rad/s), T is time (as always), and a = rad/s^2 the angular acceleration. O = ? radians is the angular distance traveled in T time. Do you see the similarities between the linear and angular equation? That holds for all of the motion equations; there is an angular equivalent for each and every linear one. So you learn the linear equations and you've automatically learned the angular ones as they are logically the same; only the units change. Your problem is a bit different as a, the angular acceleration, is not fixed, it varies with T^2. But that's no problem. O = wT + 1/2 aT^2 T^2 = 3 T^4; where w = 0, starts from rest. a(T) = aT^2 is given with a = 6 rad/s^4. So there you are O = 1/2 6T^2 T^2 = 3 T^4 radians ANS. EX: After T = 2 seconds, the wheel turns O = 3*2^4 = 48 radians. (about 8 rotations).
A wheel starts from rest. It has an angular acceleration equal to 4 rad/s^2. After it completes 10 revolutions, what will be its angular velocity?
Three of Newton’s equations of motion for LINEAR motion are shown below:[math]S=V_{i}t+\frac{1}{2}at^{2}[/math] ————--equation 1[math]V_{f}=V_{i}+at[/math] —————equation 2combine equation 1 and equation 2 to eliminate “t” gives[math]V_{f}^{2}-V_{i}^{2}=2aS[/math] —————equation 3The same three equations for ROTATIONAL motion are:[math]\theta=\omega_{i} t+\frac{1}{2}\alpha t^{2}[/math] ————--equation 4[math]\omega_{f}=\omega_{i}+\alpha t[/math] —————equation 5combine equation 4 and equation 5 to eliminate “t” gives[math]\omega_{f}^{2}-\omega_{i}^{2}=2\alpha \theta[/math] —————equation 6Be sure to use the proper units in these equations.[math]\omega[/math] = angular velocity [math][\frac{radians}{sec}][/math][math]\alpha[/math] = angular acceleration [math][\frac{radians}{s^2}][/math][math]\theta[/math] = angular displacement [math][radians][/math]It is highly recommended to watch your signs in these equations. For linear motion, velocities are up = positive, down = negative and the acceleration due to gravity always points down so [math]a=-9.81 m/s^{2}[/math]. For rotational problems, choose a direction to be positive. I like to choose clockwise = positive and counterclockwise = negative.Choose an equation or a combination of equations based on the unknown variable you are after and based on the information given in the problem. If you were trying to solve for time or if you were given time in the problem, then obviously you will use equation 1 or 2 or 4 or 5 since equations 3 and 6 are not functions of time. In your rotational problem, time is unknown, so that leaves only equation 6. So let’s use equation 6:I’ve included a conversion factor to convert [revolutions] to [radians][math]\omega_{f}^{2}-\omega_{i}^{2}=2\alpha \theta[/math][math]\omega_{f}^{2}-0=2(4 \frac{rad}{s^2})(10 rev)(2 \pi \frac{rad}{rev})[/math][math]\omega_{f}=22.4 \frac{rad}{s}[/math]Note how my units work out. If you simply used 10 revolutions for angular displacement (without the conversion factor) then the units would be all wrong.
A wheel starts rotating from rest and attains an angular velocity of 60 rad/seconds. What will the total angular displacements in radians be?
You cannot calculate angular displacement from this little information. Has it accelerated to this angular speed in 1 second? 10 seconds? or 10 years? obviously the angular displacement changes.
A wheel rotates with a constant acceleration of 2 rad/s^2. If the wheel starts from the rest, how many revolutions will it make in the first 10s?
Consider the second equation of motion:s = u*t + 0.5*a*(t^2)Since the wheel starts from rest the u*t component is zero. So, s, which in this case is the angular distance, is 100 rads, which roughly equals 15.9 full revolutions as one revolution is 2π rads. The wheel would thus make 15 revolutions in 9.7s and will be in its 16th revolution at the 10s mark.