At t=0, a flywheel has an angular velocity of 4.7 rad/s, aconstant angular accel

Question

At t=0, a flywheel has an angular velocity of 4.7 rad/s, aconstant angular acceleration of -.25 rad/s2 , and areference line at 0=0.(a) Through what maximumangle max will the reference line turn in thepositive direction? What are the (b) first and (c) second times thereference line will be at =(1/2)max?Atwhat(d) negative time and (e) positive time will the reference linebe at =-10.5 rad? (f) graph versus t, and indicate theanswers to (a) through (e) on the graph What does it mean turn in the positive direction? At t=0, a flywheel has an angular velocity of 4.7 rad/s, aconstant angular acceleration of -.25 rad/s2 , and areference line at 0=0.(a) Through what maximumangle max will the reference line turn in thepositive direction? What are the (b) first and (c) second times thereference line will be at =(1/2)max?Atwhat(d) negative time and (e) positive time will the reference linebe at =-10.5 rad? (f) graph versus t, and indicate theanswers to (a) through (e) on the graph What does it mean turn in the positive direction?

Explanation / Answer

the angular acceleration of magnitude0.25rad/s2 in the negative direction is assumed to beconstant over a large time interval , in cluding negative valuesfor t (a) we specify max with the condition = 0                                   max = -02/2 = -4.7/2(-0.25) = 44rad (b) we find values for t1 when the angulardisplacement relative its orientitation at t =0 is1 = 22rad                                       1 =0t1+1/2t12                                         t1= -0 +-02+2/ plug values we get two roots 5.5s and 32s . thus, thefirst time the reference line will be at 1 = 22radis t = 5.5s (c) the seconed time the reference line will be at1 = 22rad is t = 32s we find values for t2 when the angulardisplacement is 2 = - 10.5rad from the quadratic formula, we have 2 =0t2+1/2t22                                                            t2= - 0+_02 +22/ solving we get two roots   - 2.1sand 40s thus at t =- 2.1s the reference line will be at 2 = -10.5rad (e) at t = 40s the reference line will be at2 = - 10.5rad.                                                            t2= - 0+_02 +22/ solving we get two roots   - 2.1sand 40s thus at t =- 2.1s the reference line will be at 2 = -10.5rad (e) at t = 40s the reference line will be at2 = - 10.5rad.

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