Example: Rolling Motion Use conservation of energy Ei=U +0=Mgh Er=0+K千=1/2M2+12102 1/2MV2+1/2(1/2MR2)(vR)2 Mah= 1/2 Mv2+ 1/4 Mv2 v2=4/3gh V=(4/3gh)12 Physics 121: Lecture 19, Pg 11
Physics 121: Lecture 19, Pg 11 Example : Rolling Motion Use conservation of energy. Ei = Ui + 0 = Mgh Ef = 0 + Kf = 1/2 Mv 2 + 1/2 I 2 = 1/2 Mv 2 + 1/2 (1/2MR2 )(v/R)2 Mgh = 1/2 Mv 2 + 1/4 Mv 2 v 2 = 4/3 g h v = ( 4/3 g h ) 1/2
Lecture 19, ACT 1 Rolling Motion A race TWo cylinders are rolled down a ramp. They have the same radius but different masses, M1>M2 Which wins the race to the bottom A) Cylinder 1 B)Cylinder 2 C)It will be a tie Ⅵ2 Active Figure Physics 121: Lecture 19, Pg 12
Physics 121: Lecture 19, Pg 12 Lecture 19, ACT 1 Rolling Motion A race !! Two cylinders are rolled down a ramp. They have the same radius but different masses, M1 > M2 . Which wins the race to the bottom ? A) Cylinder 1 B) Cylinder 2 C) It will be a tie M1 h M? M2 Active Figure
Remember our roller coaster Perhaps now we can get the ball to go around the circle without anyone dying Note Radius of loop R Radius of ball r Physics 121: Lecture 19, Pg 13
Physics 121: Lecture 19, Pg 13 Remember our roller coaster. Perhaps now we can get the ball to go around the circle without anyone dying. Note: Radius of loop = R Radius of ball = r
How high do we have to start the ball Use conservation of energy Also we must remember that the minimum speed at the top is Vtop =(gR) E1=mgh+0+0 E2=mg2R+1/2m2+1/2lo2 2mgR+1/2m(gR)+1/2(2/5mr2)()2 2mgR+ 1/2 mgR+(2/10)m(gR) 2.7 mgR Physics 121: Lecture 19, Pg 14
Physics 121: Lecture 19, Pg 14 How high do we have to start the ball ? Use conservation of energy. Also, we must remember that the minimum speed at the top is vtop = (gR)1/2 E1 = mgh + 0 + 0 E2 = mg2R + 1/2 mv 2 + 1/2 I2 = 2mgR + 1/2 m(gR) + 1/2 (2/5 mr2 )(v/r)2 = 2mgR + 1/2 mgR + (2/10)m (gR) = 2.7 mgR 1 2
How high do we have to start the ball E1=mgh+0+0 E2=2.7 mgR mgh =2.7 mgR h=27R h=1.35D (The rolling motion added an extra 2/10R to the height) Physics 121: Lecture 19, Pg 15
Physics 121: Lecture 19, Pg 15 How high do we have to start the ball ? E1 = mgh + 0 + 0 E2 = 2.7 mgR mgh = 2.7 mgR h = 2.7 R h = 1.35 D (The rolling motion added an extra 2/10 R to the height) 1 2