Math Problem Statement

An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) What is its angular acceleration in rad/s2? rad/s2 (b) What is the tangential acceleration (in m/s2) of a point 8.50 cm from the axis of rotation? m/s2 (c) What is the radial acceleration in m/s2 and multiples of g of this point at full rpm? a in m/s m/s2 a as a multiple of g g

Solution

Let's solve each part step-by-step.

Given dataLet's solve each part step-by-step.

Given

  • Final angular velocity (ωf\omega_f) = 100,000 rpm
  • Time (tt) = 2.00 min
  • Radius (rr) = 8.50 cm = 0.085 m

First, let's convert angular velocity from rpm to rad/s.

ωf=100,000rpm×2πrad1rev×1min60s=10472.0rad/s\omega_f = 100,000 \, \text{rpm} \times \frac{2 \pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = 10472.0 \, \text{rad/s}

(a) Angular Acceleration

Angular acceleration α\alpha is calculated using the formula: α=ωfωit\alpha = \frac{\omega_f - \omega_i}{t} where ωi=0\omega_i = 0 (since it starts from rest) and tt in seconds is 2.00min×60s/min=120s2.00 \, \text{min} \times 60 \, \text{s/min} = 120 \, \text{s}.

α=10472.0rad/s0120s=87.27rad/s2\alpha = \frac{10472.0 \, \text{rad/s} - 0}{120 \, \text{s}} = 87.27 \, \text{rad/s}^2

Answer (a): Angular acceleration, α=87.27rad/s2\alpha = 87.27 \, \text{rad/s}^2.


(b) Tangential Acceleration

Tangential acceleration ata_t at a point on the edge of the disc is given by: at=α×ra_t = \alpha \times r where r=0.085mr = 0.085 \, \text{m} and α=87.27rad/s2\alpha = 87.27 \, \text{rad/s}^2.

at=87.27rad/s2×0.085m=7.418m/s2a_t = 87.27 \, \text{rad/s}^2 \times 0.085 \, \text{m} = 7.418 \, \text{m/s}^2

Answer (b): Tangential acceleration, at=7.418m/s2a_t = 7.418 \, \text{m/s}^2.


(c) Radial Acceleration

Radial (or centripetal) acceleration ara_r is calculated using: ar=ωf2×ra_r = \omega_f^2 \times r where ωf=10472.0rad/s\omega_f = 10472.0 \, \text{rad/s} and r=0.085mr = 0.085 \, \text{m}.

ar=(10472.0rad/s)2×0.085m=9.308×106m/s2a_r = (10472.0 \, \text{rad/s})^2 \times 0.085 \, \text{m} = 9.308 \times 10^6 \, \text{m/s}^2

To express ara_r in terms of gg, where g=9.81m/s2g = 9.81 \, \text{m/s}^2: arg=9.308×106m/s29.81m/s2=9.49×105g\frac{a_r}{g} = \frac{9.308 \times 10^6 \, \text{m/s}^2}{9.81 \, \text{m/s}^2} = 9.49 \times 10^5 \, g

Answer (c):

  • Radial acceleration ar=9.308×106m/s2a_r = 9.308 \times 10^6 \, \text{m/s}^2
  • As a multiple of gg: ar=9.49×105ga_r = 9.49 \times 10^5 \, g

Would you like further details on these calculations or have any questions?


  1. What are the differences between tangential and radial accelerations in rotational motion?
  2. How does converting rpm to rad/s impact calculations in rotational physics?
  3. How would increasing the radius impact both tangential and radial accelerations?
  4. What practical considerations limit the maximum rpm achievable in an ultracentrifuge?
  5. Why are radial accelerations often much larger than tangential accelerations at high angular speeds?

Tip: Always convert units to SI (m, s, rad) before starting calculations to maintain consistency and avoid errors.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Angular Motion
Acceleration

Formulas

Angular acceleration formula: α = (ω_f - ω_i) / t
Tangential acceleration formula: a_t = α * r
Radial acceleration formula: a_r = ω^2 * r

Theorems

Conversion of rpm to rad/s
Centripetal acceleration

Suitable Grade Level

College level - Introductory Physics