Q#16
The surface density (mass/area) of a circular disc of radius a depends on the distance from the center as ρ(r) = A + Br. Find its moment of inertia about the line perpendicular to the plane of the disc through its center.Answer:
Consider a ring of radius r from the center with a width dr. Area of the ring = 2πrdr
Mass of the ring = (A + Br)(2πr)dr
M.I. of this ring about the axis perpendicular to the plane of the disc through its center
= (A + Br)(2πr)dr(r²) = (2πr³A + 2πr⁴B)dr
M.I of the disc = ∫(2πr³A + 2πr⁴B)dr
{Limit of integration from r =0 to r = a}
= [2πr⁴A/4 + 2πr⁵B/5], putting the limits we get,
= 2π{Aa⁴/4 + Ba⁵/5}
Q#17
A particle of mass m is projected with a speed u at an angle θ with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.
Answer:
Let t be time taken to reach the highest point. At the highest point velocity in the vertical direction is = 0. Hence,
0 = u.sinθ – gt
t = usinθ/g
Horizontal distance traveled in this time
d = ucosθ(usinθ)/g
= u²sinθcosθ/g
The weight of the particle W = mg
Position vector of the particle at the highest point taking the point of projection as origin = r
The torque of the weight about the point of projection = r x W
= Weight x perpendicular distance of the line of weight from the origin.
= mgd
= mgu²sinθ(cosθ)/g
= mu²sinθcosθ
The direction of the torque will be perpendicular to the plane of motion.
Q#18
A simple pendulum of length l is pulled aside to make an angle θ with the vertical. Find the magnitude of the torque of the weight w of the bob about the point of suspension. When is torque zero?
Answer:
Taking the point of suspension as origin, position vector of the bob = r
The weight of the bob = w
The torque of the weight of the bob about the point of suspension
= r x w
= wlsinθ {Direction of the torque will be perpendicular to the plane of motion}
In the above expression, w and l are constant, θ is variable. The torque will be zero when the value of sinθ is zero. sinθ will be zero for θ = 0. Which will be when the bob is at the lowest point.
Q#19
When a force of 6.0 N is exerted at 30° to a wrench at a distance of 8 cm from the nut, it is just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut?
Answer:
Let us resolve the force of 6.0 N along the wrench and perpendicular to it.
The component of the force along the wrench, P = 6.0cos30°
Since it passes through the nut, its torque about the nut = 0
The component of the force perpendicular to the wrench, P' = 6.0 sin30° = 3.0 N
The torque of this force about the nut = 3.0(8/100) = 0.24 Nm
This much torque is required to just move the nut.
The torque of the force F about the nut = F(16/100) Nm
Equating the two we get,
F(16/100) = 0.24
F = 24/16 = 3/2 = 1.5 N
Q#20
Calculate the total torque acting on the body shown in figure (10-E2) about the point O.
Answer:
We take the torque positive which produces an anticlockwise effect.
Total torque = 20(0.04) sin30° + 5(0) + 15(0.06)sin37° - 10(0.04)sin90° Nm
= 20(0.02) + 15(0.036) – 0.40 Nm
= 0.40 + 0.54 – 0.40 Nm
= 0.54 Nm, Anticlockwise
Q#21
A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its center.
Answer:
Three forces act on the block. Weight mg, Friction F and Normal force N. The weight acts at the center of the block. Its components along the plane are mg.sinθ and perpendicular to the plane mg.cosθ. Since the block slides with uniform speed, the friction F = mg.sinθ. But these two equal forces are not in line. While mg.sinθ acts at the center C, the friction F acts on the surface. These two antiparallel forces are a/2 distance apart. So, the torque produced by them
= mg.sinθ(a/2) = ½mgasinθ. (Clockwise in the figure below).
0 = u.sinθ – gt
t = usinθ/g
Horizontal distance traveled in this time
d = ucosθ(usinθ)/g
= u²sinθcosθ/g
The weight of the particle W = mg
Position vector of the particle at the highest point taking the point of projection as origin = r
The torque of the weight about the point of projection = r x W
= Weight x perpendicular distance of the line of weight from the origin.
= mgd
= mgu²sinθ(cosθ)/g
= mu²sinθcosθ
The direction of the torque will be perpendicular to the plane of motion.
A simple pendulum of length l is pulled aside to make an angle θ with the vertical. Find the magnitude of the torque of the weight w of the bob about the point of suspension. When is torque zero?
Answer:
Taking the point of suspension as origin, position vector of the bob = r
The weight of the bob = w
The torque of the weight of the bob about the point of suspension
= r x w
= wlsinθ {Direction of the torque will be perpendicular to the plane of motion}
In the above expression, w and l are constant, θ is variable. The torque will be zero when the value of sinθ is zero. sinθ will be zero for θ = 0. Which will be when the bob is at the lowest point.
When a force of 6.0 N is exerted at 30° to a wrench at a distance of 8 cm from the nut, it is just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut?
Answer:
Let us resolve the force of 6.0 N along the wrench and perpendicular to it.
Since it passes through the nut, its torque about the nut = 0
The component of the force perpendicular to the wrench, P' = 6.0 sin30° = 3.0 N
The torque of this force about the nut = 3.0(8/100) = 0.24 Nm
This much torque is required to just move the nut.
The torque of the force F about the nut = F(16/100) Nm
Equating the two we get,
F(16/100) = 0.24
F = 24/16 = 3/2 = 1.5 N
Calculate the total torque acting on the body shown in figure (10-E2) about the point O.
We take the torque positive which produces an anticlockwise effect.
Total torque = 20(0.04) sin30° + 5(0) + 15(0.06)sin37° - 10(0.04)sin90° Nm
= 20(0.02) + 15(0.036) – 0.40 Nm
= 0.40 + 0.54 – 0.40 Nm
= 0.54 Nm, Anticlockwise
A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its center.
Answer:
Three forces act on the block. Weight mg, Friction F and Normal force N. The weight acts at the center of the block. Its components along the plane are mg.sinθ and perpendicular to the plane mg.cosθ. Since the block slides with uniform speed, the friction F = mg.sinθ. But these two equal forces are not in line. While mg.sinθ acts at the center C, the friction F acts on the surface. These two antiparallel forces are a/2 distance apart. So, the torque produced by them
= mg.sinθ(a/2) = ½mgasinθ. (Clockwise in the figure below).
Since the block does not rotate, it means there is an equal and opposite torque on the block. Consider the other force, the component of weight perpendicular to the plane mg.cosθ which acts at the center and the normal force N which does not pass through the center C but at some point down the plane. (In previous chapters we considered N to act through the center because we did not consider the torque then). Since in the perpendicular direction there is no movement, hence N = mgcosθ and they act some distance apart. The torque produced by the normal force N about the center = Torque produced by the friction F about the center = ½mga.sinθ.
Q#22
A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its center. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the center. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts.
Answer:
The torque on the rod, T = FL/4
M.I. of the rod, I= mL²/12
Angular acceleration, α = T/I = FL(12/4mL²)
= 3FL/mL²
The angle rotated by the rod during time t = θ = 0(t) + ½αt²
= ½(3FL/mL²)t²
= 3FLt²/2mL²
= 3Ft²/2mL
Q#22
A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its center. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the center. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts.
Answer:
The torque on the rod, T = FL/4
M.I. of the rod, I= mL²/12
Angular acceleration, α = T/I = FL(12/4mL²)
= 3FL/mL²
The angle rotated by the rod during time t = θ = 0(t) + ½αt²
= ½(3FL/mL²)t²
= 3FLt²/2mL²
= 3Ft²/2mL
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