Ask question

Ask question

All categories

soldier1979 [14.2K]

3 years ago

15

A string is wrapped several times around the rim of a small hoop with a radius of 8.00 cm and mass 0.180 kg. The free end of the

string is held in place and the hoop is released from rest.
After the hoop has descended 75.0 cm, calculate (a) the angular speed of the rotating hoop and (b) the speed of its center.

1 answer:

LekaFEV [45]3 years ago

8 0

Answer:

Vcm = 2.71m/s

Explanation:

From the problem description, the upper end is held fixed and not pulled upward so no work is done on the system of the hoop and string by holding this end of the string.

Mechanical energy is conserved because the string does not slip, although there is friction between the string and hoop.

So we can use the equation of conservation of mechanical energy.

K1 + U1 = K2 + U2

K1, K2 = initial and final kinetic energies of the system

U1, U2 = initial and final potential energies

Vcm = velocity of the center of mass of the system.

R = radius of the hoop

K1 = 0 system was initially at rest

U1 = mgh

K2 = 1/2MVcm² + 1/2×I×ω²

I = moment of inertia of the loop

I = MR²

ω = Vcm/R = Angukar frequency

K2 = 1/2×MVcm² + 1/2× MR² ×(Vcm/R)²

K2 = 1/2MVcm² + 1/2MVcm²

K2 = MVcm²

U2 = 0 chosen reference point as the system falls a distance of h

h = 75cm = 0.75m

Substituting the respective terms in the mechanical energy conservation equation above

0 + Mgh = MVcm² + 0

Mgh = MVcm²

gh = Vcm²

Vcm = √gh = √9.8×0.75 = 2.71m/s.

You might be interested in

Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet

Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times(5\times10^{-2})^2$

$A=7.85\times10^{-3}\ m^2$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}$

$C=6.94\times10^{-12}\ F$

We need to calculate the resistance of the wire

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}$

$R=4.27\times10^{-3}\ \Omega$

We need to calculate the voltage

Using formula of charge

$q=CV$

$V=\dfrac{q}{C}$

Put the value into the formula

$V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}$

$V=1.801\times10^{3}\ V$

(a). We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

$I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}$

$I=421779.85\ A$

$I=4.217\times10^{5}\ A$

(b). We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}$

$E=11.2\times10^{5}\ N/C$

The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2$

$E=1.126\times10^{-5}\ J$

The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Hence, (a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

8 0

3 years ago

Two small conducting point charges, separated by 0.4 m, carry a total charge of 200 C. They repel one another with a force of 12

Lunna [17]

Answer:

200 C

Explanation:

Let C1 and C2 be their charges. According to Coulomb's law

$F_C = k\frac{C_1C_2}{R^2}$

where k = $8.99\times10^9 nm^2/C^2$ is the constant, R = 0.4m is the distance between them, F = 120 N is their resulting charge force

$120 = 8.99\times10^9\frac{C_1C_2}{0.4^2}$

$C_1C_2 = \frac{120*0.4^2}{8.99\times10^9} = 2.13\times10^{-9}$

Since their total charge is 200C:

$C_1 + C_2 = 200$ or $C_1 = 200 - C_2$

We can substitute the above equation

$C_1C_2 = (200 - C_2)C_2 = 2.13\times10^{-9}$

$-C_2^2 +200C_2 - 2.13\times10^{-9} = 0$

$C= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$C= \frac{-200\pm \sqrt{(200)^2 - 4*(-1)*(-0.00000000213)}}{2*(-1)}$

$C= \frac{-200\pm200}{-2}$

$C = 1.06 \times 10^{-11}$ or $C \approx 200$

So the larger charge is C = 200 C

8 0

3 years ago

Liz rushes down onto a subway platform to find her train already departing. she stops and watches the cars go by. each car is 8.

Snezhnost [94]

The average velocity can be calculated using the formula:

v = d / t

For the 1st car, the velocity is calculated as:

v1 = 8.60 m / 1.80 s = 4.78 m / s

While that of the 2nd car is:

v2 = 8.60 m / 1.66 s = 5.18 m / s

Now we can solve for the acceleration using the formula:

v2^2 = v1^2 + 2 a d

Rewriting in terms of a:

a = (v2^2 – v1^2) / 2 d

a = (5.18^2 – 4.78^2) / (2 * 8.6)

a = 0.23 m/s

Therefore the train has a constant acceleration of about 0.23 meters per second.

5 0

3 years ago

determine the percent yield of copper by dividing the actual yield (in moles) by the theoretical yield (in moles). percent yield

Elden [556K]

Answer:

79% is = to the percent yield

3 0

3 years ago

10. According to Newton's First Lawy , if a box is pushed with no external resistance, what will the action of the box be ?

kati45 [8]

Explanation:

According to Newton's First Law of motion, if a box is pushed with no external resistance, the box will keep on moving due to the absence of external force. It might gets stopped due to frictional force that is acting between the surface and the ball. The first law of motion is also known as law of inertia. the magnitude of force acting on the object is given by second law of motion.

4 0

3 years ago

Read 2 more answers