Q#16
A ladder is resting with one end on a vertical wall and the other end on a horizontal floor. Is it more likely to slip when a man stands near the bottom or near the top?Answer:
The ladder tends to slip more when the man stands near the wall.
Let us first draw a diagram to understand the problem. AB is the ladder with end A resting on the ground and end B against the wall. OA = d, OB = h, w’ = weight of the ladder and w is the weight of the man on the ladder at distance x from end A horizontally. nA and nB are normal forces at the ends A and B also fA and fB are friction forces at A and B.
Assuming µ same at both contact surfaces,
fA = µnA, fB = µnB
Equating forces vertically,
nA + fB = w + w’
nA + µnB = w + w’
Equating forces horizontally
fA = nB
Taking Moments of all forces about B, we have
w(d – x) + w’(½d) + fAh – nAd = 0
w(d – x) + w’(½d) + µnAh – nAd = 0
w(d – x) + ½w’d = nA(d – µh)
nA = [w(d – x) + ½w’d]/(d – µh) (*)
Overturning Torque (anticlockwise) about A
= wx + w’(½d)
Restoring torque (Clockwise)
= nBh + fBd
So net overturning torque T
= wx + w’(½d) – (nBh + fBd)
= wx + w’(½d) – (nBh + µnBd)
= wx + w’(½d) – fA(h + µd)
= wx + w’(½d) – µnA(h + µd)
From (*),
= wx + w’(½d) – µ{[w(d – x) + ½w’d]/(d – µh)}(h + µd)
= wx + w’(½d) – [wµ(d – x)(h + µd) + ½µw’d(h + µd)]/(d – µh)
= wx + w’(½d) – [wµ(d – x)(h + µd) + ½µw’d(h + µd)]/(d – µh)
= wx + w’(½d) – [(wµd – wµx)(h + µd) + ½µw’dh + ½µ2w’d2)]/(d – µh)
= wx + w’(½d) – [(wµdh + wµ2d2 – wµxh – wµ2xd) + ½µw’dh + ½µ2w’d2)]/(d – µh)
= [wxd – µwxh + ½ w’d2 – ½µw’dh – wµdh – wµ2d2 + wµxh + wµ2xd – ½µw’dh – ½µ2w’d2]/(d – µh)
= [wxd + ½ w’d2 – µw’dh – wµdh – wµ2d2 + wµ2xd – ½µ2w’d2]/(d – µh)
= [wxd(1 + µ2) – wµd(µd + h) + ½ w’d(d – 2µh – ½µ2d]/(d – µh)
In this expression all except x are constant and we can see that as x increases T also increases.
So, the ladder tends to slip more when the man stands near the wall.
Q#17
When a body is weighed on an ordinary balance we demand that the arm should be horizontal if the weights on the two pans are equal. Suppose equal weights are put on the two pans, the arm is kept at an angle with the horizontal and released. Is the torque of the two weights about the middle point (Point of support) zero? Is the total torque zero? If so, why does the arm rotate and finally become horizontal?
Answer:
In fact, middle point (Point of support) is not in a straight line which joins the hanging point of the panes. This point of support (Middle point, pivot) is slightly above the center of the arms. Thus these three points make a triangle with a wide base. See diagram below,
When the arms are horizontal, the resultant weight of the pans 2w (Downard) acts in the middle and the balancing normal force 2w also acts upwards a bit above but in the same line. So net torque is zero.
When the arm is kept at an angle with the horizontal, the lines of actions of resultant weight and the normal force are not the same but are at some distance x. (See the second diagram). These two equal and opposite forces each equal to 2w but at distance d produce a net restoring torque = 2Wx which acts till the arms are horizontal because only then x = 0 i.e. the torque is zero. Now it is in stable equilibrium.
Q#18
The density of the rod AB continuously increases from A to B. Is it easier to set it in rotation by clamping it at A and applying perpendicular force at B or by clamping it at B and applying the force at A?
Answer:
By clamping at B and applying the force at A.
It is due to the fact that in this position the center of mass is comparatively closer to the clamp. Hence moment of inertia I of the rod about the clamp will be less. So, for the same angular acceleration, the torque needed (Iα) will be less.
Q#19
When tall buildings are constructed on earth, the duration of day-night slightly increases. Is it true?
Answer:
Yes, True. When tall buildings are constructed, the effective radius of the earth slightly increases. Though the mass remains constant, the moment of inertia of the earth I slightly increase. To conserve the angular momentum L of the earth the angular velocity ω = L/I slightly decrease. This means that the earth rotates slower now and takes more time to complete one rotation. So the duration of day-night slightly increases.
Q#20
If the ice at the poles melts and flows towards the equator, how will it affect the duration of day-night?
Answer:
The duration of day-night slightly increases.
The reason is same as above problem. The moment of inertia I increases.
Q#21
A hollow sphere, a solid sphere, a disc and a ring all having same mass and radius are rolled down on an inclined plane. If no slipping takes place, which one will take the smallest time to cover a given length?
Answer:
Since each object has the same radius, the total torque on each of them will be same. So angular acceleration α = Γ/I. Thus the object with minimum I will have maximum angular acceleration. Let us see the value of I for given objects:
Hollow sphere = (2/3)MR²
Solid Sphere = (2/5)MR²
Disc = ½MR²
Ring = MR²
Obviously, minimum I of them is for the solid sphere. So the solid sphere will take the smallest time to cover a given length.
Q#22
A sphere rolls on a horizontal surface. Is there any point on the sphere which has a vertical velocity?
Answer:
Theoretically, no point on the sphere has velocity vector pointing in the vertical direction.
Practically, at the points very near to the point which is in contact at the instant has a nearly vertical velocity with a very small magnitude. Just at the contact point, when the velocity becomes true vertical, its magnitude becomes zero. So, we can not say that a vertical velocity.
Assuming µ same at both contact surfaces,
fA = µnA, fB = µnB
Equating forces vertically,
nA + fB = w + w’
nA + µnB = w + w’
Equating forces horizontally
fA = nB
Taking Moments of all forces about B, we have
w(d – x) + w’(½d) + fAh – nAd = 0
w(d – x) + w’(½d) + µnAh – nAd = 0
w(d – x) + ½w’d = nA(d – µh)
nA = [w(d – x) + ½w’d]/(d – µh) (*)
Overturning Torque (anticlockwise) about A
= wx + w’(½d)
Restoring torque (Clockwise)
= nBh + fBd
So net overturning torque T
= wx + w’(½d) – (nBh + fBd)
= wx + w’(½d) – (nBh + µnBd)
= wx + w’(½d) – fA(h + µd)
= wx + w’(½d) – µnA(h + µd)
From (*),
= wx + w’(½d) – µ{[w(d – x) + ½w’d]/(d – µh)}(h + µd)
= wx + w’(½d) – [wµ(d – x)(h + µd) + ½µw’d(h + µd)]/(d – µh)
= wx + w’(½d) – [wµ(d – x)(h + µd) + ½µw’d(h + µd)]/(d – µh)
= wx + w’(½d) – [(wµd – wµx)(h + µd) + ½µw’dh + ½µ2w’d2)]/(d – µh)
= wx + w’(½d) – [(wµdh + wµ2d2 – wµxh – wµ2xd) + ½µw’dh + ½µ2w’d2)]/(d – µh)
= [wxd – µwxh + ½ w’d2 – ½µw’dh – wµdh – wµ2d2 + wµxh + wµ2xd – ½µw’dh – ½µ2w’d2]/(d – µh)
= [wxd + ½ w’d2 – µw’dh – wµdh – wµ2d2 + wµ2xd – ½µ2w’d2]/(d – µh)
= [wxd(1 + µ2) – wµd(µd + h) + ½ w’d(d – 2µh – ½µ2d]/(d – µh)
In this expression all except x are constant and we can see that as x increases T also increases.
So, the ladder tends to slip more when the man stands near the wall.
Q#17
When a body is weighed on an ordinary balance we demand that the arm should be horizontal if the weights on the two pans are equal. Suppose equal weights are put on the two pans, the arm is kept at an angle with the horizontal and released. Is the torque of the two weights about the middle point (Point of support) zero? Is the total torque zero? If so, why does the arm rotate and finally become horizontal?
Answer:
In fact, middle point (Point of support) is not in a straight line which joins the hanging point of the panes. This point of support (Middle point, pivot) is slightly above the center of the arms. Thus these three points make a triangle with a wide base. See diagram below,
When the arm is kept at an angle with the horizontal, the lines of actions of resultant weight and the normal force are not the same but are at some distance x. (See the second diagram). These two equal and opposite forces each equal to 2w but at distance d produce a net restoring torque = 2Wx which acts till the arms are horizontal because only then x = 0 i.e. the torque is zero. Now it is in stable equilibrium.
Q#18
The density of the rod AB continuously increases from A to B. Is it easier to set it in rotation by clamping it at A and applying perpendicular force at B or by clamping it at B and applying the force at A?
Answer:
By clamping at B and applying the force at A.
It is due to the fact that in this position the center of mass is comparatively closer to the clamp. Hence moment of inertia I of the rod about the clamp will be less. So, for the same angular acceleration, the torque needed (Iα) will be less.
Q#19
When tall buildings are constructed on earth, the duration of day-night slightly increases. Is it true?
Answer:
Yes, True. When tall buildings are constructed, the effective radius of the earth slightly increases. Though the mass remains constant, the moment of inertia of the earth I slightly increase. To conserve the angular momentum L of the earth the angular velocity ω = L/I slightly decrease. This means that the earth rotates slower now and takes more time to complete one rotation. So the duration of day-night slightly increases.
Q#20
If the ice at the poles melts and flows towards the equator, how will it affect the duration of day-night?
Answer:
The duration of day-night slightly increases.
The reason is same as above problem. The moment of inertia I increases.
Q#21
A hollow sphere, a solid sphere, a disc and a ring all having same mass and radius are rolled down on an inclined plane. If no slipping takes place, which one will take the smallest time to cover a given length?
Answer:
Since each object has the same radius, the total torque on each of them will be same. So angular acceleration α = Γ/I. Thus the object with minimum I will have maximum angular acceleration. Let us see the value of I for given objects:
Hollow sphere = (2/3)MR²
Solid Sphere = (2/5)MR²
Disc = ½MR²
Ring = MR²
Obviously, minimum I of them is for the solid sphere. So the solid sphere will take the smallest time to cover a given length.
Q#22
A sphere rolls on a horizontal surface. Is there any point on the sphere which has a vertical velocity?
Answer:
Theoretically, no point on the sphere has velocity vector pointing in the vertical direction.
Practically, at the points very near to the point which is in contact at the instant has a nearly vertical velocity with a very small magnitude. Just at the contact point, when the velocity becomes true vertical, its magnitude becomes zero. So, we can not say that a vertical velocity.
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