of the car when fOs is 60m/s-, the magnitude of the centripetal acceleration of the car when f500s is_3.375m/- the magnitude of the angular acceleration is_0.0lrad/s-, the magnitude of the tangential acceleration is 1.5m/s olutions ()The angular speed of the car when os 00-30 0.2 rad/ r150 (2) The angular speed 5.0 s later is a==Oo=0.15 rad/s (3)The magnitude of the centripetal acceleration of the car when fOs is 302 60m/s2 150 (4)The magnitude of the centripetal acceleration of the car when f500s is a.=02r=0.152×150=3.375m/s (5)According to the equation 4.49 @(0)=@o+at, the angular acceleration is 0.15-0.2 0.01rad/s2 150×001=1.5m/s2 ched at speed vo at an angle 0(with the horizontal) from the bottom of a hill of constant slope B as Figure 2. The Osin (0-B) FI g cos B The coordinate system is shown in figure. Assume the range is r;, the time is t, we have B os 0t sin B=vo sin At Solve the 2vo cos8sin(8-B)of the car when t=0s is 6.0m/s2 , the magnitude of the centripetal acceleration of the car when t=5.00s is 3.375m/s2 , the magnitude of the angular acceleration is 0.01rad/s2 , the magnitude of the tangential acceleration is 1.5m/s2 . Solution: (1) The angular speed of the car when t=0s is 0.2 rad/s 150 0 30 0 = = = r v ω (2) The angular speed 5.0 s later is 0.15 rad/s 4 3 ω = ω0 = (3) The magnitude of the centripetal acceleration of the car when t=0s is 2 2 2 6.0 m/s 150 30 = = = r v ac (4) The magnitude of the centripetal acceleration of the car when t=5.00s is 2 2 2 a = r = 0.15 ×150 = 3.375 m/s c ω (5) According to the equation 4.49 ω t = ω +αt 0 ( ) , the angular acceleration is 0 2 0.01 rad/s 5 ( ) 0.15 0.2 = − − = − = t ω t ω α (6) The magnitude of the tangential acceleration is 2 = α =150× 0.01 =1.5m/s τ a r . 6. A projectile is launched at speed v0 at an angle θ (with the horizontal) from the bottom of a hill of constant slope β as shown in Figure 2. The range of the projectile up the slope is β θ θ β 2 2 0 cos 2 cos sin( ) g v r − = Solution: The coordinate system is shown in figure. Assume the range is r, the time is t, we have ⎪⎩ ⎪ ⎨ ⎧ = − = 2 0 0 2 1 direction : sin sin direction : cos cos y r v t gt x r v t β θ β θ Solve the two equations, we have β θ θ β 2 2 0 cos 2 cos sin( ) g v r − = 0 v r θ β Range Fig.2 y x