You have finished running a marathon and you are relaxing and your body is

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

vredina [299]

Answer:

The angular speed after 6s is $\omega = 1466.67s^{-1}$ .

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$ , with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$ , at a distance $d = 6cm =0.06m$ , to a ring with the moment of inertia of $I =mr^2$ ; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I}$

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2}$

$\alpha = 244.44\: s^{-2}$

Therefore, the angular speed $\omega$ which is

$\omega = \alpha t$

after 6 seconds is

$\omega = 244.44$\: s^{-2}* 6s$

$\boxed{\omega = 1466.67s^{-1}}$

7 0

3 years ago

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters

Tanya [424]

Answer:

The change in potential energy of the mass as it goes up the incline is 0.343 joules.

Explanation:

We must remember in this case that change in the potential energy is entirely represented by the change in the gravitational potential energy. From Work-Energy Theorem and definition of work we get that:

$U_{g}= m\cdot g\cdot \Delta y$

Where:

$U_{g}$ - Gravitational potential energy, measured in Joules.

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Change in vertical height, measured in meters.

This work is the energy needed to counteract effects of gravity at given vertical displacement.

If we know that $m = 0.5\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.07\,m$ , the change in the potential energy of the mass as it goes up the incline is: