Spring or is used to monitor the rotational speed of a vertical rod. The sensor

Question

Spring or is used to monitor the rotational speed of a vertical rod. The sensor cousists of 6) a. A sad st js fixed to the rod and rotates with it. The angle the cone makes with the vertical There is a sunall weight, with mass m-0.500 kg that slides on the inside surface The coefficient of kinetic friction between the weight and the cone is 0.3(x) of the o efficient of static friction is .-0.400. The wealt is also attached to a spring with nstant k-7400. N m and equilibrium length 3.00cm. The spring is rig dly fixed the weight so that it always remain perfectly horizontal, and its ther end is attached to a that rides frictionlessly on the rotating rod. The speed is seused by measuring the that i b. the cone. The coeficient o spring c small loop height of the weight in the cone as it spins at the given speed. a) If the speed of the rodl is very auall but steaudily iacrasel fron rest to 1.200.rpm b) Supposing the speed of the rod was inecreased from rest to 1,200. rpm very rapidly. Describe what will be the beight of the weight in the cone? the motion of the weight in that case. Figure 6 A speed sensor consisting of a cone that is rigidly fixed to a vertically rotating rod and rotates with it, is used to monitor the rotational speed of the rod. The cone makes an angle of -30.0° with the vertical- There is a sinall weight, with mass m-0.50kg that slides on the inside surface of the cone. The coefficient of kinetic friction between the weight and the cone is 0.300 and the coefficient of static friction is ,-0400, The weight is also attached to a spring with spring constant k-7.000. N/m and equilibrium length 6- 300. The spring is rigidly fixed to the weight so that it always remain ectly horizontal, and its other end is attached to a small loop that rides frictionlessly on the rotating rod. The speed is sensed by measuring the height of the weight in the cone as it spins at the given speed

Explanation / Answer

6. given theta = 30 deg

m = 0.5 kg

uk = 0.3

us = 0.4

k = 7 N/m

lo = 3 cm

a. w = 1,200 rpm = 1200*2*pi/60 = 125.6637 rad/s

let height of the mass be m

then radius of rotation, r = h*tan(theta)

now, from force balance

Let normal reaction be N

then

Ncos(theta) + fsin(theta) + k(lo - h*tan(theta)) = mw^2*h*tan(theta)

assuming the direciton of friction downwards along the incline

also

Nsin(theta) - fcos(theta) = mg

also, for gradual motion change, we assume static firciton acts and the value of friction is the maximum possible value

then

f = us*N

hence

Ncos(theta) + us*N*sin(theta) + k(lo - h*tan(theta)) = mw^2*h*tan(theta)

Nsin(theta) - us*Ncos(theta) = mg

N = mg/(sin(theta) - us*cos(theta))

(cos(theta) + us*sin(theta))mg/(sin(theta) - us*cos(theta)) + k*lo - kh*tan(theta) = mw^2*h*tan(theta)

h = [(cos(theta) + us*sin(theta))mg/(sin(theta) - us*cos(theta)) + k*lo]/(w^2 + k)tan(theta)

putting values

h = 0.7842 mm

b. when the angular speed is increase form 1200 rpm very rapidly, the box moves too rapidly too and hence the frcition acting on it becomes kinetic friction. moreover the direction of friction becomes along the incline, but directed downwards as the box now has the tendency to move upwards due to instantanes increase in angular speed

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