PRO BLEM 10: A PARTICLE IN A PARABOLIC BOWL:= krSuppose a particle of mass m lives inside a parabolic bowl. The height z of thebowl is given, as a function of it's radius as z = kr², where k is a number. On aparticular day the particle finds itself at a height zo as measured from the bottomof the parabola. At that instant it has a horizontal speed vo. Ignoring frictioncalculate(1) for a given value of horizontal velocity, that is we will let vo → Vh, the particlewill move in a perfect circle. What is this v,? Answer this part of the question bydrawing a free body diagram and applying New ton's laws.(2) Now, suppose vo > vh, where, vh is the value you calculated above. Whatis the maximum height reached by the particle? Answer this part of the questionby evoking the conservation of energy, and remember to discuss what happens toangular momentum here.(3) Now, suppose vo = 0, but the particle is still at 20. Let us assume that 20is small and that he particle will undergo small oscillations. What is the period ofthese oscillations? You may use an energy conservation approach here. Rememberz is small, therefore # < , and that conser vation of energy implies that = 0.

(1) Consider the FBD of this problem as shown below: Here, r0 is the bowl’s radius corresponding to its height (z0). The particle at (r0, z0) experiences its weight (W) downwards, the centrifugal force (FC) due to its velocity (v0) outwards, and the Normal force (N) normal to the bowl’s surface. The forces W and FC have been resolved along and normal to the bowl’s surface with the subscripts A and N, respectively. The angle (θ) is the angle between r0 and the bowl’s surface at the particle’s position. For a very small z0, it can be approximated that the tangent of θ is z0/r0.Let g denote the gravitational acceleration. The particle does not move normal to the bowl’s surface. In case of a perfectly circular motion, the particle also does not move along the bowl’s surface. Hence, Newton’s second law at equilibrium can be applied along the bowl’s surface by taking the upward direction along the bowl to be positive and vh may be determined as follows: 0=FCA-WFCA=WAFCcosθ=Wsinθmvh2r0=mgtanθvh=gr0z0r0=gz0(2) For a v0 > vh, the particle will experience a greater FC, but the same W. As a result, its N will increase to compensate for the increased FCN so that the net force normal to the bowl’s surface stays zero. But, since the forces along the bowl’s surface are imbalanced, the particle moves in the direction of FCA, that is, up the bowl. As it travels up the bowl, the particle loses some speed and gains some radius. Hence, its FC goes on effectively decreasing. It travels up the bowl only so far that its WA and FCA become equal, and then performs circular motion at that height. Let K and P denote its kinetic and potential energies, respectively, and let all the quantities corresponding to the lower and upper positions be denoted by the subscripts 0 and 1, respectively. Since its energy is conserved, the maximum height (h1) of the particle for its speed (v1) can be determined as follows: K1+P1=K0+P012mv12+mgz1=12mv02+mgz0mv12=m12v02+gz0-z1FC1=mr112v02+gz0-z1 ∵FC1=mv12r1mgtanθ1=mkz112v02+gz0-z1 ∵z1=kr12, FC1=mgtanθ10=14v04+g2z02+v02gz0-v02g-2g2z0-kg2tan2θz1+g2z12Solve the above quadratic equation in z1 to determine the maximum height (z1 > z0) as follows: z1=--v02g-2g2z0-kg2tan2θ±-v02g-2g2z0-kg2tan2θ2-4g214v04+g2z02+v02gz02g2=v022g-k2tan2θ-z0±v022g-z0-k2tan2θ2-v022g+z02=v022g-k2tan2θ-z0±v022g-z0-k2tan2θ+v022g+z0v022g-z0-k2tan2θ-v022g+z0=v022g-k2tan2θ-z0±ktanθz0-v022g+k4tan2θ(3) Since v0 = 0 at z0, the particle gains some speed (v > v0) as it drops to the bowl’s bottom during its simple harmonic motion as z0 is small. Use the 0 subscript for the higher position and no subscript for the bottom position, determine v, and the particle’s angular velocity (ω) and period (T) corresponding to this as follows: K+P=K0+P012mv2+mg0=12m02+mgz0v=2gz0ω=2gz0 ∵v=z0ωT=π2z0g ∵T=2πω

Answered: Suppose a particle of mass m lives…