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Natasha_Volkova [10]
3 years ago
11

A 1840-kg car travels on a banked, horizontal curve of diameter 225 m. Find the maximum safe speed if the coefficient of frictio

n between the tires and the road is 0.90 and the banking angle is 6.0°. Additionally, what net force does the car experience in this case?
Physics
1 answer:
BigorU [14]3 years ago
8 0

Answer:

35 m/s

Explanation:

m = 1840 kg

2R = 225 m

R = 112.5 m

μ = 0.9

θ = 6 degree

use the formula for the banking of road

v = \sqrt{\frac{Rg(Sin\theta +\mu Cos\theta )}{Cos\theta -\mu Sin\theta }}

v = \sqrt{\frac{112.5\times 9.8(Sin6 +0.9 Cos6 )}{Cos6 -0.9Sin6 }}

v = 35 m/s

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To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

Here

n = Number of node

T = Tension

\mu = Linear density

L = Length

Replacing the values in the frequency and value of n is one for fundamental overtone

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})

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Similarly plug in 2 for n for first overtone and determine the value of frequency

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})

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Similarly plug in 3 for n for first overtone and determine the value of frequency

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

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2 years ago
A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If
vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

  • The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
  • The traslational part can be written as follows:

       K_{trans} = \frac{1}{2}* M* v_{cm} ^{2}  (1)

  • The rotational part can be expressed as follows:

       K_{rot} = \frac{1}{2}* I* \omega ^{2}  (2)

  • where I = moment of Inertia regarding the axis of rotation.
  • ω = angular speed of the rotating object.
  • If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

       v = \omega * R (3)

  • For a solid cylinder, I = M*R²/2 (4)
  • Replacing (3) and (4)  in (2), we get:

       K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2}  (5)

  • Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

       K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2}  +\frac{1}{4}* M* v_{cmc}^{2}  =  \frac{3}{4}* M* v_{cmc} ^{2} (6)

  • Repeating the same steps for the spherical shell:

        I_{sph} = \frac{2}{3} * M* R^{2} (7)  

       K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2}  (8)

      K_{sph} = \frac{1}{2}* M* v_{cms} ^{2}  +\frac{1}{3}* M* v_{cms}^{2}  =  \frac{5}{6}* M* v_{cms} ^{2} (9)

  • Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
  • And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
  • Rearranging, and taking square roots on both sides, we get:

       \frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)

  • This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.
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