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disa [49]
3 years ago
10

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Physics
1 answer:
Fantom [35]3 years ago
8 0

a) Angular acceleration: 17.0 rad/s^2

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

W_p =mg sin \theta

where \theta is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta

where L is the length of the pencil.

The relationship between torque and angular acceleration \alpha is

\tau = I \alpha (1)

where

I=\frac{1}{3}mL^2

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for \alpha, we find:

\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when \theta=10^{\circ} is

\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude mg

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

\tau = Fd

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

\tau =mg\frac{L}{2} cos \theta

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

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PLEASE PROVIDE AN EXPLANATION<br><br> THANK YOU!
Rus_ich [418]

Answer:

(a) 0.993 s

(b) 14.0 N/m

(c) -3.02 m/s

(d) -6.01 m/s²

Explanation:

(a) The block's position can be modeled as a cosine wave:

x(t) = A cos(ωt)

where A is the amplitude (in this case, 50 cm) and ω is the angular frequency.

At t = 0.200 s, x(t) = 15.0 cm.

15.0 cm = 50.0 cm cos((0.200 s) ω)

0.3 = cos((0.2 s) ω)

1.266 rad = (0.2 s) ω

ω = 6.33 rad/s

The period is:

T = (2π rad) (1 s / 6.33 rad)

T = 0.993

(b) For a spring-mass system, ω = √(k/m).  The mass of the block is 0.350 kg, so:

ω = √(k/m)

6.33 rad/s = √(k / 0.350 kg)

6.33 rad/s = √(k / 0.350 kg)

40.1 rad/s² = k / 0.350 kg

k = 14.0 N/m

(c) Energy is conserved:

EE₀ = EE + KE

½ kx₀² = ½ kx² + ½ mv²

kx₀² = kx² + mv²

(14.0 N/m) (0.50 m)² = (14.0 N/m) (0.15 m)² + (0.35 kg) v²

v = -3.02 m/s

Alternatively, we can take the derivative of our position equation:

v(t) = -Aω sin(ωt)

v = -(0.50 m) (6.33 rad/s) sin((6.33 rad/s) (0.2 s))

v = -3.02 m/s

(d) Sum of forces on the block:

∑F = ma

-kx = ma

a = -kx / m

a = -(14.0 N/m) (0.15 m) / (0.350 kg)

a = -6.01 m/s²

Alternatively, we can take the derivative of our velocity equation:

a(t) = -Aω² cos(ωt)

a = -(0.50 m) (6.33 rad/s)² cos((6.33 rad/s) (0.2 s))

a = -6.01 m/s²

6 0
3 years ago
Read 2 more answers
Please help me......
snow_tiger [21]
A. The statement is true
6 0
3 years ago
A computer is connected across a 110 V power supply. The computer dissipates 123 W of power in the form of electromagnetic radia
nexus9112 [7]

Answer:

81.3ohms

Explanation:

Resistance is known to provide opposition to the flow of electric current in an electric circuit.

Power dissipated by the computer is expressed as;

Power = current (I) × Voltage(V)

P = IV... (1)

Note that from ohms law, V = IR

I = V/R ... (2)

Substituting equation 2 into 1, we will have;

P = (V/R)×V

P = V²/R.. (3)

Given source voltage = 100V, Power dissipated = 123W

To get resistance R of the computer, we will substitute the given value into equation 3 to have

123 = 100²/R

R = 100²/123

R = 10,000/123

R = 81.3ohms

The resistance of the computer is 81.3ohms

3 0
3 years ago
Read 2 more answers
The distance between the earth and sun is 1.5 x 108 kilometers and the speed of light is 3.00 x 108 meters per second. Calculate
butalik [34]

Answer:

time = 8.3333 minutes.

Explanation:

distance between earth and sun = 1.5 * 10^{8}km

speed of light = 3* 10^{8}m/s

convert the distance unit from km to m so we can have uniform units.

distance between earth and sun = 1.5 *10^{8}*1000m

distance between earth and sun = 1.5 * 10^{11}m

speed = distance /time

time = distance / speed

time = \frac{1.5*10^{11} }{3*10^{8} }

= 0.5*10^{3}

time =500 sec

time = 500/60 minutes

time = 8.3333 minutes.

3 0
3 years ago
As an object moves, the distance it travels increases with time.<br><br> Agree<br> Disagree
Mnenie [13.5K]
Agree

Hope this helps
4 0
2 years ago
Read 2 more answers
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