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

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

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

### Given 
- Final angular velocity (\(\omega_f\)) = 100,000 rpm
- Time (\(t\)) = 2.00 min
- Radius (\(r\)) = 8.50 cm = 0.085 m

First, let's convert angular velocity from rpm to rad/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 \(\alpha\) is calculated using the formula:
\[
\alpha = \frac{\omega_f - \omega_i}{t}
\]
where \(\omega_i = 0\) (since it starts from rest) and \(t\) in seconds is \(2.00 \, \text{min} \times 60 \, \text{s/min} = 120 \, \text{s}\).

\[
\alpha = \frac{10472.0 \, \text{rad/s} - 0}{120 \, \text{s}} = 87.27 \, \text{rad/s}^2
\]

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

---

### (b) Tangential Acceleration

Tangential acceleration \(a_t\) at a point on the edge of the disc is given by:
\[
a_t = \alpha \times r
\]
where \(r = 0.085 \, \text{m}\) and \(\alpha = 87.27 \, \text{rad/s}^2\).

\[
a_t = 87.27 \, \text{rad/s}^2 \times 0.085 \, \text{m} = 7.418 \, \text{m/s}^2
\]

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

---

### (c) Radial Acceleration

Radial (or centripetal) acceleration \(a_r\) is calculated using:
\[
a_r = \omega_f^2 \times r
\]
where \(\omega_f = 10472.0 \, \text{rad/s}\) and \(r = 0.085 \, \text{m}\).

\[
a_r = (10472.0 \, \text{rad/s})^2 \times 0.085 \, \text{m} = 9.308 \times 10^6 \, \text{m/s}^2
\]

To express \(a_r\) in terms of \(g\), where \(g = 9.81 \, \text{m/s}^2\):
\[
\frac{a_r}{g} = \frac{9.308 \times 10^6 \, \text{m/s}^2}{9.81 \, \text{m/s}^2} = 9.49 \times 10^5 \, g
\]

**Answer (c):** 
- Radial acceleration \(a_r = 9.308 \times 10^6 \, \text{m/s}^2\)
- As a multiple of \(g\): \(a_r = 9.49 \times 10^5 \, g\)

---

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.

An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min.
(a) What is its angular acceleration in rad/s^2?
(b) What is the tangential acceleration (in m/s^2) of a point 8.50 cm from the axis of rotation?
(c) What is the radial acceleration in m/s^2 and multiples of g of this point at full rpm?

Explore the physics of an ultracentrifuge accelerating from rest to 100,000 rpm in 2 minutes, calculating angular, tangential, and radial accelerations for a point 8.50 cm from the rotation axis. This detailed solution covers concepts in rotational kinematics, unit conversions, and formulas for angular and radial accelerations.

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