The force diagram for cylinder and the force exerted by the string on the cylinder, by applying the translational form of Newton’s second law. The radius of the cylinder is 0.0 40 m and mass is 0. 20 kg . When the string is released, the cylinder accelerates at 2 3 g in downward direction.
The force diagram for cylinder and the force exerted by the string on the cylinder, by applying the translational form of Newton’s second law. The radius of the cylinder is 0.0 40 m and mass is 0. 20 kg . When the string is released, the cylinder accelerates at 2 3 g in downward direction.
Solution Summary: The author explains the force diagram for the cylinder and its force exerted by the string by applying Newton's second law.
The force diagram for cylinder and the force exerted by the string on the cylinder, by applying the translational form of Newton’s second law. The radius of the cylinder is 0.040 m and mass is 0.20 kg. When the string is released, the cylinder accelerates at 23g in downward direction.
(b)
To determine
The rotational inertia of the solid cylinder. The radius of the cylinder is 0.040 m and mass is 0.20 kg. When the string is released, the cylinder accelerates at 23g in downward direction.
(c)
To determine
The rotational acceleration of the cylinder on applying the rotational form of Newton’s second law. The radius of the cylinder is 0.040 m and mass is 0.20 kg. When the string is released, the cylinder accelerates at 23g in downward direction.
(d)
To determine
To explain: Whether the answer in part (c), is consistent with the application of a=rα. Here, a and α relate the translational acceleration and rotational acceleration. The radius of the cylinder is 0.040 m and mass is 0.20 kg. When the string is released, the cylinder accelerates at 23g in downward direction.
A small postage stamp is placed in front of a concave mirror (radius = 1.1 m), such that the
image distance equals the object distance. (a) What is the object distance? (b) What is the
magnification of the mirror (with the proper sign)?
Calculate the anti-clockwise torque and the clockwise torque of the system with the ruler and the washers. Record these values in Data Table 5. Ruler = 11.56 g, small washer = 1.85 g, large washer = 24.30 g.
Calculate the % Difference in the Torques and record the values in Data Table 5.
Is ΣAnticlockwise torque and Anticlockwise torque the same thing, are they solved in the same way?
A window washer stands on a uniform plank of mass M = 142 kg and length l = 2.80 m supported by 2 ropes attached at the ends of the plank. The window washer has a mass m = 68.0 kg. What is the tension in each of the ropes, T1 and T2, if the window washer's displacement from the center of mass of the plank is x = 0.930 m as shown in Figure 1: Window Washer Problem?
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