Q#54
A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makes an angle of 37° with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.Answer:
Let us draw a figure as below,
Length of the string OA = 50 cm = r (radius)
When the string makes an angle 37° with the vertical, the ball's height above its lowest position h= OB – OC = 50 – 50cos37° = 10 cm = 0.10 m
Mass of the ball m = 100 g = 0.10 kg
Taking the ball's lowest position as base, P.E. given to the ball = mgh (i)
At the lowest position P.E. = 0 and K.E. = ½mv², (where v = velocity)
Total Energy = ½mv² (ii)
Total energy is conserved hence (i) and (ii) will be equal.
½mv² = mgh
v² = 2gh
Let the tension in the string at its lowest position be T. Centripetal forces at this position = T – mg
Centripetal acceleration = v²/r
Hence
T – mg = mv²/r
T = mg + mv²/r
= mg + m(2gh)/r
= mg(1 + 2h/r)
= 0.10(9.8)[1 + 2(0.10)/0.50] ≈ 1.4 N
Q#55
Figure (8-E15) shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring of spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal.
Answer:
Let the required compression of the spring be x.
P.E. stored in the spring = ½kx²
When the spring is released, the spring does a work on the block and changes its K.E. which is equal to the P.E. stored in the spring = ½kx². When the block reaches P, a part of this K.E. is converted to P.E. Let v be the velocity of the block at P.
It's energy at P = K.E. + P.E.
= ½mv² + mgR (Its height above flat surface = R)
Hence, ½mv² + mgR = ½kx²
mv² + 2mgR = kx²
Since the block presses the track with a force mg, that means Normal force on the block is also mg which due to its velocity should be mv²/R. Hence,
mv²/R = mg
mv² = mgR
Putting this value in (i), we get,
mgR + 2mgR = kx²
kx² = 3mgR
x² = 3mgR/k
x = √(3mgR)/k
The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of √(3gl). Find the angle rotated by the string before it comes slack.
Answer:
Assuming l = length of the string. Speed given to the bob = u = √(3gl)
K.E. given to the bob
= ½mu² = ½m(3gl) = 3mgl/2 (i)
The string will become slack at the point where the tension in the string T = 0. Suppose this point comes after rotating an angle θ and bob's speed here is v. If at this position the string makes an angle ⲫ with the upward vertical then ⲫ = 180°- θ. Its height from the original position A is = h. (See the figure below.)
h = AC + DC = l + lcosⲫ = l(1 + cosⲫ)
Taking the level of original position of the bob as zero P.E. level, P.E. of the bob at final position
= mg(l + lcosⲫ)
= mgl(1 + cosⲫ)
And K.E. = ½mv²
Total mechanical energy at this point = ½mv² + mgl(1 + cosⲫ)
Since total mechanical energy will be conserved, hence (i) and (ii) will be equal.
Equating (i) and (ii) we get,
½mv² + mgl(1 + cosⲫ) = 3mgl/2
v² + 2gl(1 + cosⲫ) = 3gl (iii)
Consider the forces at the final position. Force along the string towards center = T+mg.cosⲫ and centripetal acceleration mv²/l. So,
T + mgcosⲫ = mv²/l
mv²/l = mgcosⲫ (since here T = 0)
v² = glcosⲫ
Putting this value in (iii)
glcosⲫ + 2gl(1 + cosⲫ) = 3gl
glcosⲫ + 2glcosⲫ = gl
3glcosⲫ = gl
3cosⲫ = 1
cosⲫ = 1/3
cos(180°- θ) = 1/3
–cosθ = 1/3
θ = cos⁻¹(–1/3)
Q#57
A heavy particle is suspended by a 1.5 m long string. It is given a horizontal velocity of √57 m/s. (a) Find the angle made by the string with the upward vertical, when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take g = 10 m/s².
Answer:
Let us draw the figure for this problem.
(a) u= √57 m/s, K.E. = ½mu² = 57m/2 J
As in the problem 56, at the point of string being slack,
mv²/l = mgcosⲫ
v² = lg cosⲫ
Total energy at final point = K.E. + P.E.
= ½mv² + mgl(1 + cosⲫ)
= ½mlgcosⲫ + mgl(1 + cosⲫ) = 57m/2
(Since final energy will be equal to the initial energy)
lgcosⲫ + 2gl(1 + cosⲫ) = 57
3glcosⲫ = 57 – 2gl
3(10)(1.5)cosⲫ = 57 – 2(10)1.5 = 27
45cosⲫ = 27
cosⲫ = 0.60 = cos 53°
ⲫ = 53°
(b) We have
v² = lgcosⲫ = 1.5(10)(3/5) = 9
v = 3.0 m/s
(c) Once the string becomes slack, the bob will move like a projectile with initial velocity = 3.0 m/s and angle of projection ⲫ = 53°. For this projectile motion maximum height above the point of projection d = (vsinⲫ)²/2g.
Let the maximum height reached above the point of suspension = x.
x = lcosⲫ + d
= lcosⲫ + (vsinⲫ)²/2g
= 1.5(3/5) + (3 x 0.8)²/2(10)
x = 1.188
x ≈ 1.2 m
Taking the level of original position of the bob as zero P.E. level, P.E. of the bob at final position
= mg(l + lcosⲫ)
= mgl(1 + cosⲫ)
And K.E. = ½mv²
Total mechanical energy at this point = ½mv² + mgl(1 + cosⲫ)
Since total mechanical energy will be conserved, hence (i) and (ii) will be equal.
Equating (i) and (ii) we get,
½mv² + mgl(1 + cosⲫ) = 3mgl/2
v² + 2gl(1 + cosⲫ) = 3gl (iii)
Consider the forces at the final position. Force along the string towards center = T+mg.cosⲫ and centripetal acceleration mv²/l. So,
T + mgcosⲫ = mv²/l
mv²/l = mgcosⲫ (since here T = 0)
v² = glcosⲫ
Putting this value in (iii)
glcosⲫ + 2gl(1 + cosⲫ) = 3gl
glcosⲫ + 2glcosⲫ = gl
3glcosⲫ = gl
3cosⲫ = 1
cosⲫ = 1/3
cos(180°- θ) = 1/3
–cosθ = 1/3
θ = cos⁻¹(–1/3)
Q#57
A heavy particle is suspended by a 1.5 m long string. It is given a horizontal velocity of √57 m/s. (a) Find the angle made by the string with the upward vertical, when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take g = 10 m/s².
Answer:
Let us draw the figure for this problem.
(a) u= √57 m/s, K.E. = ½mu² = 57m/2 J
As in the problem 56, at the point of string being slack,
mv²/l = mgcosⲫ
v² = lg cosⲫ
Total energy at final point = K.E. + P.E.
= ½mv² + mgl(1 + cosⲫ)
= ½mlgcosⲫ + mgl(1 + cosⲫ) = 57m/2
lgcosⲫ + 2gl(1 + cosⲫ) = 57
3glcosⲫ = 57 – 2gl
3(10)(1.5)cosⲫ = 57 – 2(10)1.5 = 27
45cosⲫ = 27
cosⲫ = 0.60 = cos 53°
ⲫ = 53°
(b) We have
v² = lgcosⲫ = 1.5(10)(3/5) = 9
v = 3.0 m/s
(c) Once the string becomes slack, the bob will move like a projectile with initial velocity = 3.0 m/s and angle of projection ⲫ = 53°. For this projectile motion maximum height above the point of projection d = (vsinⲫ)²/2g.
Let the maximum height reached above the point of suspension = x.
x = lcosⲫ + d
= lcosⲫ + (vsinⲫ)²/2g
= 1.5(3/5) + (3 x 0.8)²/2(10)
x = 1.188
x ≈ 1.2 m
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