PHYS 1: Assignment 4 Solutions Due on Feb 4 013 Prof. Oh Katharine Hyatt 1
Katharine Hyatt PHYS 1 (Prof. Oh ): Assignment 4 Solutions H + R 1.43 We see immediately that there is friction in this problem, so we can t use conservation of energy. We know that for the hoop: v h =ωrrolling condition And for the box: a b =g cos θ µ k g sin θ The fact that the hoop keeps pace with the box means that v h = v b Which automatically gives us a h = a b We see that: a h = d dt (ωr) = ωr = τ I h R = τ m h R R = τ m h R We realise that gravity exerts no torque because it is exerted at the centre of mass of the hoop. The only force which exerts a torque is friction. τ =fr =µ k m h g sin θr a h = µ km h g sin θr m h R =µ k g sin θ µ k g sin θ =g cos θ µ k g sin θ µ k sin θ = cos θ µ k = 1 tan θ Page of 8
Katharine Hyatt PHYS 1 (Prof. Oh ): Assignment 4 Solutions H + R 1.43 H + R 1.49 We see that there is no friction in this problem, so we can use conservation of energy. Let s call the speed of the ball as it exits the track v f : P E =mgh KE = 1 mv + I ω E =P E + KE de dt =0 mgh =mgh + 1 mv f + mr 5 ω f From the rolling condition, we see that So: ω f = v f R mgh =mgh + 1 mv f + m 5 v f g(h h) = 7 v f = 10 v f 10g(H h) 7 Then, using our kinematics knowledge, we see that: h = gt h t = g x =v f t 0(H h)h = 7 0(60 0)(0) = 7 40 =0m 7 =47.8m Page 3 of 8
Katharine Hyatt PHYS 1 (Prof. Oh ): Assignment 4 Solutions H + R 1.49 H + R 1.50 Again there is no friction, so we can use conservation of energy. In order for the marble to just stay on the track, the normal force exerted on it by the track at the top of the loop should be exactly 0. Since the marble is on a circular section of track, this is the same as saying that gravity provides all the centripetal acceleration. This condition can be written: So, from conservation of energy: g = v top R H + R 1.51 mgh =mgr + 1 mv top + I ω =mgr + 1 mr vtop mgr + 5 R = 3 mgr + 1 5 mgr = 17 10 mgr h = 17 10 R To solve this, we can just write down the torques and use Newton s third law. We know that: To find the angular acceleration α: ma =W T τ =T R =Iα = mr T R = mra = RW 3 T R =RW T = W 3 a R α = T R I = T R mr = W 3mR = g 3R RT Page 4 of 8
Katharine Hyatt PHYS 1 (Prof. Oh ): Assignment 4 Solutions H + R 1.51 H + R 1.5 We use Newton s third law again: ma =mg sin θ T where T is the tension T R =Iα I = m R T = m Rα = m R a R = m a ma =mg sin θ 1 ma 3 ma =mg sin θ a = 3 g sin θ So the tape unwinds with constant acceleration. Now we can use a kinematics equation: l =v 0 t + 1 at L = 1 at where T is the unwinding time T = L a L 3 T = g sin θ 3L = g sin θ Page 5 of 8
Katharine Hyatt PHYS 1 (Prof. Oh ): Assignment 4 Solutions H + R 1.5 H + R 1.57 We can use energy methods again: mgh = 1 Iω + 1 mv mgh = 1 v mr 5 R + 1 mv = 7 10 mv 10 v = 7 gh x =v cos θt y = v sin θt 1 gt gx y = x tan θ v cos θ 7x y = x tan θ 0h cos θ Plugging in the numbers the question gives us, we find: x = 5.34 m Page 6 of 8
Katharine Hyatt PHYS 1 (Prof. Oh ): Assignment 4 Solutions H + R 1.57 H + R 1.59 Let s call the velocity of the centre of mass v c, and the velocity of the far end (the one her hand isn t on) v f. We see that she throws the stick with ω 0. Let s call the angular velocity about the centre of mass ω. v c = ω 0L v f =ω 0 L = ω =ω 0 πn =θ(t ) =ωt v c =gt t = v c g πn =ω v c g ω 0 L gπn =ω 0 gπn =ω0l gπn ω 0 = L ay =(vc f ) (vc) i gh =v c 0 look at the falling behavior ωl h = v c g = ω 0L 4 1 g = gπnl 8Lg = πnl 4 Page 7 of 8
Katharine Hyatt PHYS 1 (Prof. Oh ): Assignment 4 Solutions H + R 1.59 H + R 1.60 We realize that the ball stops slipping when the rolling condition is met, when ωr = v We see that the only torque on the ball comes from friction: And the ball is decelerating linearly due to friction: τ =Iα fr = mr α 5 α = 5µmgR mr = 5µg R ω =αt = 5µg R v =v 0 µmgt ωr =v 0 µgt 5µgT =v 0 µgt 7µgT =v 0 T = v 0 7µg We can check our units - ms 1 m 1 s = s so our units make sense. K + K 6.35 As the block rocks sideways without slipping, the point of contact between it and the sphere is Rθ. As long as the centre of mass is inside the Rθ envelope (closer to the vertical than Rθ) then the block won t fall off. If the centre of mass obeys this condition, a restoring torque exists. If the centre of mass doesn t obey the condition, a restorting torque is impossible. Assuming that R L, the block will be unstable if: Rθ < L tan θ R <L < L θ for small angles Page 8 of 8