Answer True, False, or Cannot tell to each of the four statements below. E.g., i
ID: 1471806 • Letter: A
Question
Answer True, False, or Cannot tell to each of the four statements below. E.g., if the answer to the first statement is 'true' and to the rest, 'cannot tell', enter TCCC. A square plate can rotate about an axle through its centre. One of three forces of equal magnitude can be applied to the plate, as shown. The plate will not rotate if F is applied. A wheel turns on a frictionless axle. A string wrapped around the run of the wheel is connected to a mass. The mass is released at t = 0 and hits the floor at t = t_1. The graph shows a possible omega - t behaviour for this situation. The solid cylinders and the cylindrical shells have the same mass, the same radius, and turn on frictionless axles. A rope is wrapped around each of them and is acted on as shown. Cylinder 2 has a larger angular acceleration than cylinder shell 3. Force F_3 exerts a torque about the axle.Explanation / Answer
Part (a)
Solution: The square plate will rotate when the torque due to the applied force acts on it.
Torque is given by,
= F*r*sin
where r is the perpendicular distance between the axel and the point where the force F is applied. is the angle between the r and the F which is zero in in the case of F1. Thus the toque is,
= F1*r*sin 0o
= F1*r*0
= 0 N.m
Since torque due to the force F1 is zero, the plate will not rotate. Hence the given statement is TRUE.
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Part (b)
Let the radius of the wheel be r. When the mass m is released then the weight of the mass F = m*g acts on the wheel at a perpendicular distance r and the angle between F and r is = 90o . Thus the torque is,
= F*r*sin 90o
= F*r*1
= F*r N.m
Due to this torque, the wheel starts rotating and it undergoes the angular acceleration .
The at moment t = t1 the mass hits the floor hence no weight F acts thus no torque. However the wheel has gained the angular velocity and thus the angular momentum L. Due to the angular momentum L, the wheel will continue to rotate with angular velocity (as the wheel turns on the frictionless axel).
The graph do not show the given situation correctly hence it is WRONG.
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Part (c) The cylinder and cylindrical shell have same mass m and same radius r. But the rotational inertia of the cylinder is Ic = (1/2)mr2 while rotational inertia of the cylindrical shell Is = mr2. Hence Is > Ic
The same torque is acting on the all the cylinders and the cylindrical shells due to the same force (weights and applied force are the same).
We have,
= I* thus,
= /I
Since the inertia Ic of the cylinder is less than the inertia Is of the cylindrical shell; I for cylindrical shell is twice than I for cylinder.
is same for all, but I is smaller for cylinder. Thus the ratio /I will be larger for cylinder, that is is large.
The cylinder will have larger angular acceleration .
Thus the cylinder 2 has a larger angular acceleration than the cylindrical shell 3.
Hence the statement is TRUE.
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Part (d)
The angle between the force F3 and perpendicular distance between the line joining axel and the point where the force F3 acts is zero.
Thus the torque = F3*r*sin 0o
= 0 N.m
Thus the force F3 exerts no torque about the axel. Hence the statement is FALSE.
(Kindly note that the image is blurred and in this part it seems to me that given is Force F3 . If it is F1 then the = 90o and force exerts a torque about axel and hence statement is TRUE.)
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