physics hw question. using circular motion?
956315
The simplest way is to cut it into five slices, each with an arc of 72 degrees (or 2*pi/5 radians). You could probably also figure out a way to cut it into strips so that each strip has equal area, but that would be significantly more complicated.
716372
1 and 2 are easy. I'll help with #3.\n\n3. You need I, the rotational inertia. From a table in my physics book, I = MR^2 for the reel. You need to convert the force to torque, T = F*R. \n\na. Then you can get the angular acceleration, alpha, from \nT = I*alpha\n\nb. In t = 0.25 s, omega will reach alpha*t. The angular displacement, theta, will be (1/2)*omega*t. From that and the circumference you can get the line drawn from the reel.
235638
Remember the formula for centripetal acceleration,\na_c = v^2/r\n\nso if r is doubled but v remains constant, than the centripetal acceleration is halfed.\n\nTo prove this, lets put in some numbers.\nSay v = 2 m/s, r = 1 meter, thus, a_c = 4 m/s^2.\nNow double r to be 2 meters, the new a_c = 2 m/s^2, or 1/2 the original value.
755334
Analyze the forces at the top of the loop:\n\nThe force of gravity pulling the ball downward,\nthe upward centrifigal force keeping the ball on the track,\nand the force of the track pushing the ball downward.\n\nThe answer to the question is when the centrifigal force is exactly equal to the weight, so the force of the track equals zero.\nmg=mw^2R\nw is the angular velocity\nw=sqrt(g/R)\n\nOnce you know the angular velocity, compute the equivalent linear velocity by mutlipying by R\n\nthen use mgh=1/2mv^2\nto compute h\n\nWhere is point Q?\n\nj
914856
OK as the ball is swung around in a horizontal circle the scale inside the tube is measuring the tension on the string. By knowing the time to complete one revolution  linear velocity can be found from v=r*omega(angular velocity). in the second part the v is given and needs to be converted into radians/second by using 3.7=.5*omega where omega=3.7/5=.74rad/s then using the mass radius and angular velocity you can find the tension in the string. With a= acceleration (into center) omega^2*r=.74^2*.5=.2738m/s^2and using f=ma the tension in the string will equal f which equals .2*.2738=.0548newtons Unless i have a silly math mistake somewhere.
754995
Nicole I think that to understand what is to be done here I suggest to assume that a coordinate system be attached to one of the objects. I-don’t-remember-who says, but << any reference system at rest or moving with linear constant speed has no privilege with others the like>>.\n(x) m2=4kg is reference system #2, where v2=0, and m1=2kg is moving with speed v=8+12=20 ms, impact by m1 on m2 M1=m1*v=40kg*m/s\n(y) m1=2kg is reference system #1, where v1=0, and m2=4kg is moving with speed v=8+12=20 ms, impact by m2 on m1 M2=m2*v=80kg*m/s, so R=40/80=0.5
1067758
In physics, circular motion is rotation along a circle: a circular path or a circular orbit. The rotation around a fixed axis of a three-dimensional body involves circular motion of its parts. We can talk about circular motion of an object if we ignore its size, so that we have the motion of a point mass in a plane.
799229
a. Calculate the horizontal component of the 485 N.\nFh = 485 N*cos35\nCalculate the force of friction.\nFf = 0.57*319 N\nCalculate the net force.\nFn = Fh-Ff\nCalculate the mass of the box.\nW = m*g\nCalculate the acceleration.\nFn = m*a\nCalculate time to move s = 4.00 meters\ns = (1/2)*a*t^2\n\nb. Repeat with uk = 0.75
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1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
684454
Let's consider the case when the Car is right in front.\n\nAt that instance, the car is travelling tangentially to a circle of radius 132ft, where the Camera is the centre of the circle.\n\nCircumference of the Circle = 2*(pi)*R\n= 2 * pi * 132 = 264*(pi) ft.\n\nTime taken to travel the entire circumference by the car\nT = circumference/velocity = 264*(pi)/264 = pi seconds\n\nHence the car would require pi s to cover 360 degrees of the circle.360 degrees = 2(pi) radians.\n\nW = angular velocity = 2(pi)/pi = 2 radians/sec.\n\ncan't help about half a sec later now
850378
Centripetal acceleration a=r*w^2, r=0.35 m, w=2500*2pi/60 rad/s is angular speed, v=w*r is linear speed;
724756
For a body in circular motion there will be a centripetal force\n= M V V /R.\n\nIn this case the tension of the string provides this force.\n\nMaxinmum tension = 50N.\n\nThat is the maximum centripetal force is 50 N\n \nM V V /R = 50\n\nV V = 50 x R / M = 50 x 1/ 2 =25 (m/s)^2.\n\n The maximum speed is V = 5 m/s.
785816
h = 4tanθ\nVh = dh/dt = (dh/dθ)(dθ/dt)\ndh/dθ = 4/cos^2θ\ncos^2θ = 4^2/(4^2 + 19^2)\nVh = (16 + 361)/4)(5/361)\nVh = (5/4)(16/361 + 1)\nVh = 1.3054 km/sec
784822
According to Newton's first law of motion, a moving body travels along a straight path with constant speed (i.e., has constant velocity) unless it is acted on by an outside force.\n\nFor circular motion to occur there must be a constant force acting on a body, pushing it toward the centre of the circular path. This force is the centripetal (“centre-seeking”) force. The centrifugal (centre-fleeing) force will, therefore, be the opposing force which pushes away from the centre\n\nThe two forces are equal in magnitude and opposite in direction.
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs
797003
ω = rotational speed = v/r = 3.6/.2 = 18 r/s\nt = √(2y/a) = √(2*2.1/9.8) = .654 sec\n\nθ = ωt = 18*.654 = 11.783 rad
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is