A disc of radius R = 0.57 m and of mass M = 1.44 kg is rotating with the initial angular momentum ωi = 0.42 rev/s on a frictionless horizontal plane. You drop a small clay piece of mass m = 0.18 kg vertically, it lands, and is stuck on a point of the rim of the disc at the distance R = 0.57 m from the center of the disc (Fig.9). The initial moment of inertia of the disk about the axis of rotation through its center is Ii = （MR^2）/2. What is the moment of inertial of the disc-clay system after the clay’s landing If in kg∙m2? Keep three significant figures for the answer. Note the clay piece has negligible size so that it can be treated as a point.
A disc of radius R = 0.57 m and of mass M = 1.44 kg is rotating with the initial angular momentum ωi = 0.42 rev/s on a frictionless horizontal plane. You drop a small clay piece of mass m = 0.18 kg vertically, it lands, and is stuck on a point of the rim of the disc at the distance R = 0.57 m from the center of the disc (Fig.9). The initial moment of inertia of the disk about the axis of rotation through its center is Ii = （MR^2）/2. What is the moment of inertial of the disc-clay system after the clay’s landing If in kg∙m2? Keep three significant figures for the answer. Note the clay piece has negligible size so that it can be treated as a point.
A disc of radius R = 0.57 m and of mass M = 1.44 kg is rotating with the initial angular momentum ωi = 0.42 rev/s on a frictionless horizontal plane. You drop a small clay piece of mass m = 0.18 kg vertically, it lands, and is stuck on a point of the rim of the disc at the distance R = 0.57 m from the center of the disc (Fig.9). The initial moment of inertia of the disk about the axis of rotation through its center is Ii = （MR^2）/2. What is the moment of inertial of the disc-clay system after the clay’s landing If in kg∙m2? Keep three significant figures for the answer. Note the clay piece has negligible size so that it can be treated as a point.
A disc of radius R = 0.57 m and of mass M = 1.44 kg is rotating with the initial angular momentum ωi = 0.42 rev/s on a frictionless horizontal plane. You drop a small clay piece of mass m = 0.18 kg vertically, it lands, and is stuck on a point of the rim of the disc at the distance R = 0.57 m from the center of the disc (Fig.9). The initial moment of inertia of the disk about the axis of rotation through its center is Ii = （MR^2）/2. What is the moment of inertial of the disc-clay system after the clay’s landing If in kg∙m2? Keep three significant figures for the answer. Note the clay piece has negligible size so that it can be treated as a point.
Transcribed Image Text:Small clay
piece
Rotating disc
R
Fig. 9
Definition Definition Product of the moment of inertia and angular velocity of the rotating body: (L) = Iω Angular momentum is a vector quantity, and it has both magnitude and direction. The magnitude of angular momentum is represented by the length of the vector, and the direction is the same as the direction of angular velocity.
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