1) The blades in a blender rotate at a rate of 6540 rpm. When the motor is turne
ID: 2119328 • Letter: 1
Question
1) The blades in a blender rotate at a rate of 6540 rpm. When the motor is turned off during operation, the blades slow to rest in 3.66 s. What is the angular acceleration as the blades slow down? 2) A grinding wheel 0.500 m in diameter rotates at 2024 rpm. Calculate its angular velocity in rad/s. . What are the linear speed and acceleration of a point on the edge of the grinding wheel. Enter linear speed first. 3) How fast (in rpm) must a centrifuge rotate if a particle 7.01 cm from the axis of rotation is to experience an acceleration of 84000g's? 4) An automobile engine slows down from 4580 rpm to 1140 rpm in 2.55 s. Calculate its angular acceleration, assumed constant. Calculate the total number of revolutions the engine makes in this time. 5) A wheel 30.7 cm in diameter accelerates uniformly from 246 rpm to 372 rpm in 6.02 s. How far will a point on the edge of the wheel have traveled in this time?Explanation / Answer
1) here intial angular velocity (u) = 6540 rpm = 6540 * 2*pi / 60 rad / sec = 684.8672 rad / sec time ( t) = 3.66 sec final angular velocity (v) = 0 so angular acceleration = (v - u ) / t = ( 0 - 684.8672 )/3.66 = -187.122 rad / sec2 2) angular velocity (u) = 2024 rpm = 2024 * 2*pi / 60 rad / sec = 211.9528 rad / sec linear speed of the particle at the edge = angular velocity ( in rad / sec ) * distance from center = ( 211.9528) * ( 0.5 / 2 ) = 52.9882 m/sec 3) Let us first calculate the angular velocity in rad / sec .. Let the angular velocty be w rad / sec distance r = 7.01 cm = 0.0701 m accelarion = 84000 g = 84000 * 9.8 = 823200 m/sev accelaration = angular velocity^2 * distance 823200 = w^2 * 0.0701 so angular velocity w = 3426.8388 rad / sec so angualr velocity in RPM = 3426.8388 * 60 / (2*pi ) = 32723.9 RPM 4) intial angular velocity (u) = 4580 RPM = 4580 * 2*pi / 60 rad / sec = 479.6165 rad / sec final angualr velocity ( v ) = 1140 RPM = 1140 * 2*pi / 60 = 119.3805 rad /sec time (t) = 2.55 sec so angular accleraion (a) = (v-u) / t = (119.3805 - 479.6165) / 2.55 = -141.26902 rad / sec2 Let the angle rotated be s .. so.. s = u*t + 0.5*a*t^2 = (479.6165*2.55) + (0.5*(-141.26902) * 2.55^2) = 763.7212 radians ... so no. of revolution = angle / (2*pi ) = 763.7212 / ( 2*pi ) = 60.775 revoluitons 5) intial angular velocity (u) = 246 RPM = 246 * 2*pi / 60 rad / sec = 25.76106 rad / sec final angualr velocity ( v ) = 372 RPM = 372 * 2*pi / 60 = 38.95575 rad /sec time (t) = 6.02 sec so angular accleraion (a) = (v-u) / t = (38.95575 - 25.76106) / 6.02 = 2.191809 rad / sec2 Let the angle rotated be s .. so.. s = u*t + 0.5*a*t^2 = (25.76106*6.02) + (0.5*2.191809 * 6.02^2) = 194.7976 radians ... so distance travelled by edge = angle * distance from center = 194.7976*0.307/2 = 29.9014 metres
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