Study Guide Chapter 5 Sections 3 and 4

 If the speed changes while an object is traveling in a circle it has two types of acceleration: centripetal - due to direction change – toward center of the circle tangential – due to speed change – tangent to the circle total accel = vector sum of a T + a C total force = vector sum of F T + F c

A small sphere of mass m is attached to the end of a cord of length R, which rotates under the influence of the gravitational force in a vertical circle about a fixed point O. Let us determine the tension in the cord at any instant when the speed of the sphere is v and the cord makes an angle Θ with the vertical.

Case 1 R α vapprox true for objects that fall through the liquid at low speed or very small objects in air ex – sphere in honey, dust in air R = bv So if Σ F = ma mg – R = ma mg – bv = m(dv/dt) dv/dt = g – (b/m)v

Case 1 cont. dv/dt = g – (b/m)v Note: when a = dv/dt = 0 g = b/mv mg/b = v v is the terminal speed object no longer accelerates since R = W

Case 1 cont. A solution for our differential equation is v = mg/b (1 – e -bt/m ) Since v T = mg/b v = v T (1 – e -t/ Τ ) and T = m/b - time constant

Time constant – time it takes for 1 – e -t/T to become equal to 1 – e -1 = or the time for v = 63.2%v T

A small sphere of mass 2 g is released from rest in a large vessel filled with oil. The sphere approaches a terminal speed of 5 cm/s. Determine (a) the time constant and (b) the time it takes the sphere to reach 90% of its terminal speed.

Case 2 R α v 2 true for large objects at high speed ex – airplane, skydiver, baseball R = ½ D ρ Av 2 ρ = density of air A = cross sectional area of object measured in a plane perpendicular to velocity D = drag coeff (0.5 for spherical object in air)

Case 2 cont. If Σ F = ma mg – R = ma mg – ½ D ρ Av 2 = m(dv/dt) dv/dt = g – (D ρ A/2m)v 2

Case 2 cont. dv/dt = g – (D ρ A/2m)v 2 Note: when a = dv/dt=0g = (D ρ A/2m)v 2 v = (2mg/D ρ A) 1/2 v = terminal speed Note: R increases with v 2 R increases as ρ increases