The magnitude of the normal acceleration is A) proportional to radius of curvature.

B) inversely proportional to radius of curvature.

C) sometimes negative.

D) zero when velocity is constant.

Skip to content# The magnitude of the normal acceleration isA) proportional to radius of curvature. B) inversely proportional

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The magnitude of the normal acceleration is A) proportional to radius of curvature.

B) inversely proportional to radius of curvature.

C) sometimes negative.

D) zero when velocity is constant.

a)-It has normal acceleration only

Explanation:

When a particle moves in a curved path with uniform velocity:

-The magnitude of the speed remains constant

-The speed direction varies.

Tangential acceleration :Is defined as the change in magnitude of the tangential velocity with respect to time.

For this case, as the magnitude of the velocity is constant then the tangential acceleration does not exist.

Normal or centripetal acceleration: occurs due to the change of direction of the tangential velocity , For this case, as the direction of the velocity is varying then the normal acceleration exist, and is calculated as follows:

[tex]a_{n} =\frac{v^{2} }{r}[/tex]

[tex]a_{n} : normal acceleration[/tex]

[tex]v=tangential velocity[/tex]

r= radius of the circular path

b. inversely proportional to radius of curvature

Explanation:

In curvilinear motions, the normal acceleration which is also called the centripetal acceleration is always directed towards the center of the circular path of motion. This acceleration has a magnitude that is directly proportional to the square of the speed of the body undergoing the motion and inversely proportional to the radius of the curvature of the motion path. The centripetal or normal acceleration a, can be given by;

a = [tex]\frac{v^2}{r}[/tex]

Where;

v = speed of the body

r = radius of curvature.

I went to everybody's favorite search engine to get some data to answer this question with. I can't name it, but it rhymes with Floogle. Anyway, I searched for the phrase "zero to 60" just to get an idea of what's hot.

Floogle pointed me to a website that provided this data on fast cars:

#1). 2016 AMZ Grimsel Electric Race Car 0-60 mph 1.5 sec

#2). 2014 Porsche 918 Spyder 0-60 mph 2.3 sec

#3). 2015 Tesla Model S P90D 'Ludicrous Speed' Upgrade 0-60 mph 2.6 sec

And my favorite, because I own one:

#4). 2000 Dodge Intrepid ES 0-60 mph 8.4 sec

Here are the accelerations for the zero-to-60 times listed for these 4 cars:

#1). 17.88 m/s² (1.82 G's)

#2). 11.66 m/s² (1.19 G's)

#3). 10.32 m/s² (1.05 G's)

#4). 3.19 m/s² (0.33 of a G)

That's what these "fast cars" will do, on the track,

with their pedals mashed to the metal.

"Normal" accelerations are all less than these.

Radius of the circle will be 2.5 m

Explanation:

We have given velocity of particle moving in the circle v = 5 m/sec

Acceleration of particle in the circle [tex]a=10m/sec^2[/tex]

We have to find the radius of the circle

We know that acceleration is given by [tex]a=\frac{v^2}{r}[/tex]

So [tex]10=\frac{5^2}{r}[/tex]

[tex]r=\frac{25}{10}=2.5m[/tex]

So radius of the circle will be 2.5 m

The normal acceleration in the United States is zero to sixty time. The average rate on ordinary cars was between 3 and 4 m/s2

s = 1989.9 m

Explanation:

In order to find the distance at which the engineer should apply brake, we can use 2nd equation of motion. The second equation of motion is as follows:

s = Vi t + (0.5)at²

where,

s = distance from the station = ?

Vi = Initial velocity = 44 m/s

t = time required to stop = 27 s

a = acceleration of train = 2.2 m/s²

Therefore,

s = (44 m/s)(27 s) + (0.5)(2.2 m/s²)(27 s)²

s = 1188 m + 801.9 m

s = 1989.9 m

(1) At t = 2.0 s tangential acceleration, a₁ = 4.8 m/s² and normal acceleration , a₂ = 230.4 m/s².

(2)15.8°

(3)0.55 s

Explanation:

θ = 2 + 4t³

(1) the tangential acceleration a₁ = rα where α = angular acceleration = d²θ/dt² evaluated at t = 2.0 and r = 0.10 m

α = d²θ/dt² = 24t at t = 2 s

α = 24t = 24 × 2 = 48 rad/s²

a₁ = rα = 24tr = 0.10 m × 48 rad/s² = 4.8 m/s²

The normal acceleration a₂ = rω² where ω = angular velocity = dθ/dt = 12t² at t = 2.0 s

ω = 12t² = 12 × 2² = 48 rad/s

a₂ = rω² = 144rt⁴ = 0.1 × 48² = 230.4 m/s²

(2) The total acceleration a = √a₁² + a₂² = √[(24tr)² + (12t⁴r)²] = √[576t²r² + 144t⁸r⁴]. Since a₁ = a/2 ⇒ 24tr = √[576t²r² + 144t⁸r²]/2

48tr = √[576t²r² + 144t⁸r⁴]

squaring both sides

2304t²r² = 576t²r² + 144t⁸r²

2304t²r² - 576t²r² = 144t⁸r²

1728t²r² = 144t⁸r²

1728t²/144 = t⁸

t⁸ = 12t²

t⁸ - 12t² = 0

t²(t⁶ - 12) = 0

t² = 0 or t⁶ - 12 = 0

t = 0 or t⁶ = 12

[tex]t = \sqrt[6]{12}\\ t = 1.51 s[/tex]

The tangential acceleration equals half the total acceleration at t = 1.51 s.

θ = 2 + 4t³ = θ = 2 + 4(1.51)³ = 15.77° ≅ 15.8°

(3) We find when a₁ = a₂

24tr = 144rt⁴

24t = 144t⁴

t⁴ = t/6

t⁴ - t/6 = 0

t(t³ - 1/6) = 0

t = 0 or t³ - 1/6 = 0

t = 0 or t = 1/∛6 = 0.55 s

So the tangential acceleration equals the normal acceleration at t = 0.55 s

a) 1725.93 rpm

b) α = 3.012 rad/s²

Explanation:

The centrifugal acceleration is given by

a = v²/r

a = 1000 g = 1000 × 9.8 = 9800 m/s²

v = velocity = ?

r = radius of motion = 300 mm = 0.3 m

9800 = v²/0.3

v² = 2940

v = 54.22 m/s

v = rw

w = angular speed

54.22 = 0.3 × w

w = 54.22/0.3 = 180.74 rad/s

w = 2πf

f = frequency in rev/s

180.74 = 2πf

f = 28.77 rev/s

f = 1725.93 rpm

b) Using the equations of motion,

Since the centrifuge starts from rest,

w₀ = 0 rad/s

w = 180.74 rad/s (equivalent to the f = 1725.93 rpm)

t = 1 min = 60 s

α = ?

w = w₀ + αt

180.74 = 0 + 60α

α = 3.012 rad/s²