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xz_007 [3.2K]
4 years ago
13

At the presentation ceremony, a championship bowler is presented a 1.60-kg trophy which he holds at arm's length, a distance of

0.655 m from his shoulder joint.
(a) Determine the torque the trophy exerts about the shoulder joint when his arm is horizontal. (Enter the magnitude only.)

(b) Determine the torque the trophy exerts about the shoulder joint when his arm is at an angle of 15.0
Physics
1 answer:
nikdorinn [45]4 years ago
8 0

Answer:

a)10.28 Nm

b)9.93 Nm

Explanation:

Let g = 9.81m/s2. First we can calculate the weight of the trophy

W = mg = 1.6 * 9.81 = 15.696 N

(a) The torque is product of force and its moment arm

T = WL = 15.696 * 0.655 = 10.28 Nm

(b) Suppose his arm makes an angle of 15 degree with respect to the horizontal line. We can still calculate the arm length, or the horizontal distance from the trophy to the champion:

L_2 = Lcos(15^0) = 0.655cos(15^0) = 0.655*0.966 = 0.633 m

Again, torque is product of force and its moment arm

T_2 = WL_2 = 15.696 * 0.633 = 9.93 Nm

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude
matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, v_t = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

v_t = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, v_t = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4)  The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

4 0
3 years ago
In 1913 Niels Bohr formulated a method of calculating the different energy levels of the hydrogen atom.
Svetlanka [38]
They are right the answer is A true
4 0
3 years ago
Explain what is happening when roller coaster is at each point. **The roller coaster has started at A and goes to D.
yaroslaw [1]

Answer:

At point A, the cart has high potential energy. At point b, the cart is pulled down by gravity. At point c, the cart gains its highest kinetic energy. At point d, the cart returns back to the same state but with lower potential energy.

3 0
2 years ago
Eric has a mass of 19.0 kg on the earth. What is Eric's weight on earth? What is Eric's weight on Mars? where the acceleration o
Tom [10]

Answer:

Weight on Earth = We = 186.2 N

Weight on Mars = Wm = 70.94 N

Explanation:

The weight of an object is defined as the force applied on the object by the gravitational field. The magnitude of weight is given by the following formula:

W = mg

were,

W= Weight of Eric

m = mass of Eric

g = acceleration due to gravity

ON EARTH:

W = We = Eric's Weight on Earth = ?

m = Eric's Mass on Earth = 19 kg

ge = acceleration due to gravity on Earth = 9.8 m/s²

Therefore,

We = (19 kg)(9.8 m/s²)

<u>We = 186.2 N</u>

<u></u>

ON MARS:

W = Wm = Eric's Weight on Mars = ?

m = Eric's Mass on Mars = 19 kg

gm = acceleration due to gravity on Mars = 0.381(ge) = (0.381)9.8 m/s² = 3.733 m/s²

Therefore,

Wm = (19 kg)(3.733 m/s²)

<u>Wm = 70.94 N</u>

3 0
3 years ago
Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part
Fantom [35]

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

8 0
3 years ago
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