Given the position vector r = 3i+4j+5k and force\'s F = 16i-20j-5k, find the mag

Question

Given the position vector r = 3i+4j+5k and force's F = 16i-20j-5k, find the magnitude of the torque vector tau so that tau = r times F A solid disk of mass equal to 7.0 kg and radius of 30 cm changes its rotations gradually from 45 rpm to 120 rpm in 5.0 sec. find the angular acceleration in rad/s^2 how many revolutions has the disk turned in this 5.0 seconds? find the change in rotational kinetic energy. find its moment of inertia in kgm^2 find the magnitude of torque needed to make such a change in rpm. find the tangential acceleration of a point P on the outer rim of the disk. The fan blade in a jet engine rotates from rest with the angular acceleration alpha = 20t^2 find the angular velocity as a function of time. find the angular displacement as a function of time. if the rotational inertia of the blade is 5.0 kgm^2, find the work done on the blade in the first initial 3.0 seconds. A hollow sphere (M = 3.0 kg and R = 60 cm) is rolling down an inclined surface with a height of 2.0 meters and an inclined angle of 25 degrees.

Explanation / Answer

2)

torque = r x F

torque = (3i + 4j + 5k ) x ( 16i - 20 j - 5k)

torque = 80i + 95 j - 124 k

magnitude = sqrt(80^2 + 95^2 + 124^2) = 175.5 Nm

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3)

initial angular speed wi = 45 rpm = 45*2pi/60 = 4.712 rad/s

final angular speed wf = 120 rpm = 120*2pi/60 = 12.6 rad/s

time interval t = 5 s

angular acceleraation alpha = (wf-wi)/t = (12.6-4.712)/5 = 1.58 rad/s^2

(4)

theta = (wi + wf)*t/2

theta = (4.712 + 12.6)*5/2 = 43.28 rad = 43.28/(2pi) = 6.88 revolutions

(5)

dK = (1/2)*I*(wf^2 - wi^2)

I = (1/2)*m*R^2

dK = (1/2)*(1/2)*7*0.3^2*(12.6^2-4.712^2) = 21.5 J

(6)

moment of inertia I = (1/2)*m*R^2 = (1/2)*7*0.3^2 = 0.315 kg m^2

(7)

torque = I*alpha = 0.315*1.58 = 0.4977 Nm

(8)

atan = R*alpha = 0.3*4.712 = 1.4136 m/s^2

(9)

w = integra alpha*dt

w = 20t^3/3

(10)

angular displacement theta = integral w*dt

theta = 20t^4/12

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