Ask question

Ask question

All categories

2 months ago

9

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

22 N is applied tangentially to a sprocket of radius 6 cm for 6 s, what angular speed does the wheel achieve, assuming it rolls without slipping?

1 answer:

kicyunya [3.2K]2 months ago

7 0

Answer:

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

Explanation:

The formula

$I\alpha = Fd$

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

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

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

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

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

Thus, the angular speed $\omega$ which after 6 seconds is

$\omega = \alpha t$

You might be interested in

Complete the paragraph to describe the relationship between kinetic energy and braking distance. Use . A car moves at a speed of

Softa [3030]

ke prop to v^2

ke1/v1^2=ke2/v2^2

400/50x50=joules/100x100

400x2x2

1600j

7 0

2 months ago

Read 2 more answers

A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

kicyunya [3294]

Answer:

a) Q = C*emf

b) Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let $V_{R}$ represent the Voltage across the Resistor

$V_{c}$ and

represent the Voltage across the capacitor

Implementing KVL;

$emf - V_{R} - V_{c} = 0\\$

.........................(1)

Since this is a series connection, the same current traverses through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Integrating $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$

Substituting the values of t and I₀ into the aforementioned formula for I

$I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}$

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

$Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf$

4 0

2 months ago