Question

a) Consider a simple pendulum of mass m mounted inside a railroad car that is accelerating
to the right with constant acceleration a. Analyse this problem in the non inertial frame
of reference to find the angle alfa(F) with the vertical direction at which the pendulum will
remain at rest relative to the moving car.
b) On Jupiter a day lasts for 9.9 earth hours and the circumference at the equator is
448600 km. If the measured value of gravitational acceleration at the equator is
24.6ms^-2, what is the true gravitational acceleration and the centrifugal
acceleration.

Answer 1

Answer on Question #39666, Physics, Other

a) Consider a simple pendulum of mass $m$ mounted inside a railroad car that is accelerating to the right with constant acceleration $a$ . Analyse this problem in the non-inertial frame of reference to find the angle alfa(F) with the vertical direction at which the pendulum will remain at rest relative to the moving car.

Solution:

The non-inertial frame is the frame of reference which is not inertial or is moving.

Free body diagram of pendulum

The pendulum is at rest. The forces on the pendulum are: weight of the pendulum, mg, tension T in the string and pseudo force ma.

$$\sum F _ {x} = T \sin \theta - m a = 0$$ $$\sum F _ {y} = T \cos \theta - m g = 0$$ $$T \sin \theta = m a$$ $$T \cos \theta = m g$$

Combining two equations, we have :

$$\tan \theta = \frac {a}{g}$$ $$\theta = \tan^ {- 1} \left(\frac {a}{g}\right)$$

b) On Jupiter a day lasts for 9.9 earth hours and the circumference at the equator is $448600\mathrm{km}$ . If the measured value of gravitational acceleration at the equator is $24.6\mathrm{ms}^{\wedge}-2$ , what is the true gravitational acceleration and the centrifugal acceleration.

Solution:

Along the Jupiter equator, the outward centrifugal force is vertical and upward. Thus, it directly subtracts from the gravitational force, and the measured gravitational acceleration results from the difference:

$$g _ {\text {m e a s u r e d}} = g _ {\text {t r u e}} - g _ {\text {c e n t r i f u g a l}} = 2 4. 6 \mathrm {m} / \mathrm {s} ^ {2}$$

The angular velocity

$$\omega = \frac {2 \pi}{T} = \frac {2 \cdot 3 . 1 4}{9 . 9 \mathrm {h o u r s}} = \frac {2 \cdot 3 . 1 4}{9 . 9 \cdot 3 6 0 0 \mathrm {s}} = 0. 0 0 0 1 7 6 2 = 1 7. 6 2 \cdot 1 0 ^ {- 5} \mathrm {s} ^ {- 1}$$

The radius of Jupiter

$$R = \frac {L _ {e q u a t o r}}{2 \pi} = \frac {4 4 8 6 0 0 \mathrm {k m}}{2 \cdot 3 . 1 4} = 7 1 4 3 3. 1 2 1 \mathrm {k m}$$

The centrifugal acceleration (perceived gravity):

$$g _ {c e n t r i f u g a l} = \omega^ {2} R = (1 7. 6 2 \cdot 1 0 ^ {- 5}) ^ {2} \cdot 7 1 4 3 3. 1 \cdot 1 0 ^ {3} = 2. 2 1 7 7 = 2. 2 2 \mathrm {m / s ^ {2}}$$

The true gravitational acceleration:

$$g _ {t r u e} = g _ {m e a s u r e d} + g _ {c e n t r i f u g a l} = 2 4. 6 + 2. 2 2 = 2 6. 8 2 \mathrm {m / s ^ {2}}$$

Answer. a) $\theta = \tan^{-1}\left(\frac{a}{g}\right)$

b) $g_{\text{centrifugal}} = 2.22 \, \text{m/s}^2$ , $g_{\text{true}} = 26.82 \, \text{m/s}^2$ .