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a year ago

Physics System of Particles and Rotational Motion

Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

let M and R be mass and radius of half disc , mass per unit area of half disc

So m = $\frac{M}{\frac{1}{2} π R^{2}}$

(a) the half disc be supposed to be consist of a large number of semicircular ring of mass dm and thickness dr and radii ranging from r=0 to r=R.

surface area of semicircular ring of radius r and thickness dr = $\frac{1}{2} 2 π r d r = π r d r$

so mass of the elementary ring dm = $π r d r * \frac{2 M}{π R^{2}}$

dm= $\frac{2 M}{R^{2}} r d r$

if x,y are coordinates of centre of mass of this element,

then (x,y)=(0,2r/ $π$ )

so x=0 and y =2r/ $π$

let x_cm and y_cm be the coordinates of the centre of mass of the semicircular disc

so x_cm= $\frac{1}{M} \int_{o}^{R} x d m = \frac{1}{M} \int_{o}^{R} 0 d m = 0$

Y_cm= $\frac{1}{M} \int_{o}^{R} y d m = \frac{1}{M} \int_{0}^{R} \frac{2 r}{π} * (\frac{2 M}{R^{2}} r d r)$

= $\frac{4}{π R^{2}} \int_{0}^{R} r^{2} d r = \frac{4}{π R^{2}} [\frac{r^{3}}{3}]$ ₀^R

= 4R/3 $π$

Centre of mass of a uniform quar

...more

This is a long answer type question as classified in NCERT Exemplar

let M and R be mass and radius of half disc , mass per unit area of half disc

So m = $\frac{M}{\frac{1}{2} π R^{2}}$

(a) the half disc be supposed to be consist of a large number of semicircular ring of mass dm and thickness dr and radii ranging from r=0 to r=R.

surface area of semicircular ring of radius r and thickness dr = $\frac{1}{2} 2 π r d r = π r d r$

so mass of the elementary ring dm = $π r d r * \frac{2 M}{π R^{2}}$

dm= $\frac{2 M}{R^{2}} r d r$

if x,y are coordinates of centre of mass of this element,

then (x,y)=(0,2r/ $π$ )

so x=0 and y =2r/ $π$

let x_cm and y_cm be the coordinates of the centre of mass of the semicircular disc

so x_cm= $\frac{1}{M} \int_{o}^{R} x d m = \frac{1}{M} \int_{o}^{R} 0 d m = 0$

Y_cm= $\frac{1}{M} \int_{o}^{R} y d m = \frac{1}{M} \int_{0}^{R} \frac{2 r}{π} * (\frac{2 M}{R^{2}} r d r)$

= $\frac{4}{π R^{2}} \int_{0}^{R} r^{2} d r = \frac{4}{π R^{2}} [\frac{r^{3}}{3}]$ ₀^R

= 4R/3 $π$

Centre of mass of a uniform quarter disc

Mass per unit area of the quarter disc = $\frac{M}{\frac{π R^{2}}{4}} = \frac{4 M}{π R^{2}}$

For half disc along yaxis and xaxis =4R/3 $π$

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New question posted

a year ago

Physics Semiconductor Electronics

Q.6. Let $A = [- 1, 1]$ , Then, discuss whether the following functions defined on $A$ are one-one, onto, or bijective:

(i) $f (x) = \frac{x}{2}$

(ii) $g (x) = | x |$

(iii) $h (x) = x | x |$

(iv) $k (x) = x^{2}$ .

0 Follower 10 Views

New question posted

a year ago

Physics Semiconductor Electronics

Q.6. Let $A = [- 1, 1]$ , Then, discuss whether the following functions defined on $A$ are one-one, onto, or bijective:

(i) $f (x) = \frac{x}{2}$

(ii) $g (x) = | x |$

(iii) $h (x) = x | x |$

(iv) $k (x) = x^{2}$ .

0 Follower 7 Views

New question posted

a year ago

Physics Systems of Particles and Rotational Motion

Q.6. Let $A = [- 1, 1]$ , Then, discuss whether the following functions defined on $A$ are one-one, onto, or bijective:

(i) $f (x) = \frac{x}{2}$

(ii) $g (x) = | x |$

(iii) $h (x) = x | x |$

(iv) $k (x) = x^{2}$ .

0 Follower 4 Views

New question posted

a year ago

Physics Systems of Particles and Rotational Motion

Q.6. Let $A = [- 1, 1]$ , Then, discuss whether the following functions defined on $A$ are one-one, onto, or bijective:

(i) $f (x) = \frac{x}{2}$

(ii) $g (x) = | x |$

(iii) $h (x) = x | x |$

(iv) $k (x) = x^{2}$ .

0 Follower 3 Views

New answer posted

a year ago

Physics Motion in a Plane

4.32

(a) Show that for a projectile the angle between the velocity and the x-axis as a function

of time is given by

(b) Shows that the projection angle _0 for a projectile launched from the origin is

given by

where the symbols have their usual meaning.

0 Follower 47 Views

Vishal Baghel

Contributor-Level 10

4.32

(a) Let $θ (t) = \tan^{- 1} ? (\frac{v_{o y} - g t}{v_{o x}})$ be the angle at which the projectile is fired w.r.t. the x-axis, since $θ_{0} = \tan^{- 1} ? (\frac{4 h_{m}}{R})$ depends on t

Therefore tan $θ$ since (Vy=Voy -gt and Vx = Vox)

(b) Since $θ$ = $θ (t) = \frac{V x}{V y} = \frac{V o y - g t}{V o x}$ sin2 $θ (t) = \tan^{- 1} ? (\frac{V o y - g t}{V o x})$ /2g…….(1)

R = $h_{m a x}$ sin2 $u^{2}$ /g …….(2)

Dividing (1) by (2)

( $θ$ /R) = [ $u^{2}$ sin2 $θ$ /2g]/[ $h_{m a x}$ sin2 $u^{2}$ /g] = $θ$ / 4

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New answer posted

a year ago

Physics Motion in a Plane

4.31 A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

0 Follower 80 Views

Vishal Baghel

Contributor-Level 10

4.31

Speed of the cyclist = 27 km/h = 7.5 m/s

Radius of the road = 80m

The net acceleration is due to braking and the centripetal acceleration

Acceleration due to braking = 0.5 m/s²

Centripetal acceleration a = v²r = (7.5)2/80= 0.703 m/s²

The resultant acceleration is given by a = sqrt ( $a_{t}^{2}$ + $a_{c}^{2}$ ) = sqrt ( ${0.5}^{2}$ + ${0.7}^{2}$ ) = 0.86 m/s²

tan $β$ = AC / at = 0.7/0.5, $β$ = 54.5 $°$

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