(2)According to o(oxr(t=v( and r(r. o(xr(o=0, we know r(0)v(0=0 that is v(t is always perpendicular to r(o) 3. The angle turned through by the flywheel of a generator during a time interval I is given by 0=at+br'-cr, where a, b, and c are constant. Its angular velocity is a+3b12-4cr.And its angular acceleration is 6bt-12ct Solution de Its angular velocity is a at+br'-ct*)=a+ 3bt-4ct And its angular acceleration is B-do d(a+ 3br2-4c1)=6bt-12ct2 4. The flywheel of an engine I rotating at 25.2 rad/. When the engine is turned off, the flywheel decelerates at a constant rate and comes to rest after 197s. The angular acceleration(in rad/s2)of the flywheel is 12.8 rad/s2, the angle(in rad) through which the flywheel rotates in coming to rest is 248. 1 rad, and the number of revolutions made by the flywheel in coming to rest is_ 39.5 Solution: (1)According to the equation 4.49 @(0=@o+at, the angular acceleration is 25.2 =-128rad/ (2) According to the equation 0()=0+o+a, the change angle is △=t+-at2=252×197-×128×19.72=248rad (3)The number of revolutions is \.1 39.5 2x2×3.14 5 A car is traveling around a banked, circular curve of radius 150 m on a test track. At the instant when f=Os, the car is moving north, and its speed is 30.0 m/s but decreasing uniformly, so that after 5.0 s its angular speed will be 3/4 that it was when FOs. The angular speed of the car when fOs is 0.2rad/s, the angular speed 5.0 s later is _ 0. 15rad/s, the magnitude of the centripetal acceleration(2) According to (t) r(t) v(t) r r v ω × = and r(t)⋅[ (t)× r(t)] = 0 r r r ω , we know r(t)⋅ v(t) = 0 r v , that is v(t) r is always perpendicular to r(t) r . 3. The angle turned through by the flywheel of a generator during a time interval t is given by 3 4 θ = at + bt − ct , where a, b, and c are constant. Its angular velocity is 2 3 a + 3bt − 4ct . And its angular acceleration is 2 6bt −12ct . Solution: Its angular velocity is 3 4 2 3 ( ) 3 4 d d d d at bt ct a bt ct t t = = + − = + − θ ω And its angular acceleration is 2 3 2 ( 3 4 ) 6 12 d d d d a bt ct bt ct t t = = + − = − ω β 4. The flywheel of an engine I rotating at 25.2 rad/s. When the engine is turned off, the flywheel decelerates at a constant rate and comes to rest after 19.7s. The angular acceleration (in rad/s2 ) of the flywheel is 12.8 rad/s2 , the angle (in rad) through which the flywheel rotates in coming to rest is 248.1 rad , and the number of revolutions made by the flywheel in coming to rest is 39.5 . Solution: (1) According to the equation 4.49 ω t = ω +αt 0 ( ) , the angular acceleration is 0 2 1.28 rad/s 19.7 25.2 = − = − = − t ω α (2) According to the equation 2 0 2 1 (t) t t θ = θ +ωo + α , the change angle is 1.28 19.7 248.1 rad 2 1 25.2 19.7 2 1 2 2 ∆θ = ωot + αt = × − × × = (3) The number of revolutions is 39.5 2 3.14 248.1 2 = × = ∆ π θ 5 A car is traveling around a banked, circular curve of radius 150 m on a test track. At the instant when t=0s, the car is moving north, and its speed is 30.0 m/s but decreasing uniformly, so that after 5.0 s its angular speed will be 3/4 that it was when t=0s. The angular speed of the car when t=0s is 0.2rad/s , the angular speed 5.0 s later is 0.15rad/s , the magnitude of the centripetal acceleration