Title ROTATIONAL DYNAMICS - GCE A - L FURTHER MATHEMATICS-1_070929 (1).pdf ✅

ROTATIONAL DYNAMICS COMPILED BY KENG ELSON MN. www.tefaafrik.com 1 0775 FURTHER MATHEMATICS [email protected] ROTATIONAL DYNAMICS INTRODUCTION Dynamics involves studying the motion of a body by taking into consideration the forces acting on the body. It is the opposite to kinematics which mainly studies the motion of the body with no regards to the forces involved. Rotational dynamics mainly focuses on the motion of rotating bodies. In general, every rotating body must possess an axis of rotation, a velocity and thus kinetic energy, mechanical energy, a moment of inertia, an angular momentum and many more. MOMENT OF INERTIA The moment of inertia of a body is the measure of the body’s resistance to change in its angular velocity. Considering a body (point mass) B of mass mB rotating about an axis (Δ), distant r0 from it as shown in below Mathematically, the moment of inertia I of bofy B about the axis (Δ) is the product of the mass (mB) of body B and the square of the distance (r0 ) of the body from the axis ⇒ I = mBr0 2 . The moment of inertia I is measured in kgm2 Example: Find the moment of inertia of a point mass of 7kg rotating at a distance of 4cm from an axis Solution I = mr 2 = 7 ( 4 100) 2 kgm2 = 7 ( 1 25) 2 kgm2 = 7 625 kgm2 a. MOMENT OF INERTIA OF A LARGE BODY I (Δ) mB B r0 ω (Δ) r0 r1 r2 m0 m1 m2 B
ROTATIONAL DYNAMICS COMPILED BY KENG ELSON MN. www.tefaafrik.com 2 0775 FURTHER MATHEMATICS [email protected] The large body B is composed of non-identical point masses of masses mi each a distance of ri from a given axis of rotation (Δ) where i = 0, 1, 2, 3, ... n. By definition, the overall moment of inertia of body B about the axis of rotation (Δ) is the sum of the moment of inertia’s of its point masses about the axis of rotation (Δ) i.e. is the sum of the product of its composed non-identical masses and the square of their respective distances from the axis of rotation ⇒ I = I0 + I1 + I2 + ⋯ + In = m0r0 2 + m1r1 2 + m2r2 2 + ⋯ + mnrn 2 ⇒ I = ∑miri 2 n i=0 Example: Find the moment of inertia of a system consisting of 2kg, 4kg and 5kg masses rotating about an axis (Δ1 ) and distances of 2m, 3m and 8m from the axis of rotation Solution: I = I0 + I1 + I2 = ∑miri 2 2 i=0 = m0r0 2 + m1r1 2 + m2r2 2 ∶ { m0 = 2kg m1 = 4kg m2 = 5kg and { r0 = 2m r1 = 3m r2 = 4m ⇒ I = 2(2) 2kgm2 + 4(3) 2kgm2 + 5(4) 2kgm2 = 8kgm2 + 48kgm2 + 80kgm2 ⇒ I = 136kgm2 Note: If the body has a total mass of M and all the point masses are equidistant from the axis of rotation (Δ) then ri = r = cst ⇒ I = ∑miri 2 n i=0 = (∑mi n i=0 ) r 2 ∶ But (∑mi n i=0 ) = M ∴ I = Mr 2 This is a special case of a ring of mass M and radius r where all point masses are on the circumference and rotate about the center of the ring b. RADIUS OF GYRATION, k This is the length that represents the distance in a rotating system between the point about which it is rotating and the point to or from which a transfer of energy has the same maximum effect. In other words, the radius of gyration K is the distance between the axis of rotation and the line that passes through the center of gravity G and parallel to the axis of rotation of any rotating object. (Δ) 2m 3m 5m 2kg 4kg 5kg
ROTATIONAL DYNAMICS COMPILED BY KENG ELSON MN. www.tefaafrik.com 3 0775 FURTHER MATHEMATICS [email protected] KINETIC ENERGY OF A ROTATING BODY Consider a point mass m rotating about a distance r from an axis (Δ1 ) K. E = 1 2 mv 2 : v = ωr ⇒ K. E = 1 2 m(ωr) 2 = 1 2 mω 2 r 2 = 1 2 ω 2 (mr 2 ) ∶ But I = mr 2 ⇒ K. E = 1 2 Iω 2 MOMENT OF INERTIA BY INTEGRATION There are various steps to follow when finding the moment of inertia of a uniform body using integration. The useful steps are listed below STEP 1: Make a sketch diagram showing axis of rotation (or hinge) and indicate the Cartesian axis of circular axis at the hinge STEP 2: Divide the body into several parts called increments of mass δm and length δl at a distance of r from the axis of rotation STEP 3: Define a constant λ as the mass per unit length, area or volume of the body. The choice between these three parameters is the largest for the body on question STEP 4: Evaluate the mass δm of an increment in terms of λ and define its moment of inertia as δI = δm × r 2 in terms of λ STEP 5: Move to total derivative i.e. δI to dI and as well δl to dl by considering the entire body STEP 6: Integrate both sides with limiting values to get the moment of inertia I of the whole body Axis of rotation G Radius of gyration, k I = ∑miri 2 n i=0 = Mk 2 ⇒ K = ඨ I M TABLE FAN NOTE (Δ1 ) m r ω
ROTATIONAL DYNAMICS COMPILED BY KENG ELSON MN. www.tefaafrik.com 4 0775 FURTHER MATHEMATICS [email protected] Note: Do not forget to replace λ by its expression a. Moment of inertia of a Uniform rod of mass m and length 2a about an axis through its center and perpendicular it - Let λ be the mass per unit length of the uniform rod ⇒ λ = m 2a - The moment of inertia of the increment about the axis is δI = δm × l 2 - The mass per unit length of the increment λ = δm δl ⇒ δm = λ × δl ∴ δI = λl 2δl and note that −a ≤ l ≤ a - For the total rod, we take the total derivative ⇒ dI = λl 2dl ⇒ ∫ dI = λ ∫ l 2dl a −a ⇒ I = 2λ ∫ l 2dl a 0 = 2λ [ l 3 3 ] 0 a = 2λ ( a 3 3 ) λ = m 2a ⇒ I = 2 ( m 2a ) ( a 3 3 ) = ma 2 3 ∶ Hence the moment of inertia of a uniform rod of mass m and length 2a rotating about an axis through its center is thus I = ma 2 3 b. Moment of inertia of a Uniform rod of mass m and length 2a about an axis through one of its ends and perpendicular it - Let λ be the mass per unit length of the uniform rod ⇒ λ = m 2a - The moment of inertia of the increment about the axis is δI = δm × l 2 - The mass per unit length of the increment λ = δm δl ⇒ δm = λ × δl ∴ δI = λl 2δl and note that 0 ≤ l ≤ 2a - For the total rod, we take the total derivative ⇒ dI = λl 2dl ⇒ ∫ dI = ∫ λl 2dl 2a 0 ⇒ I = λ ∫ l 2dl 2a 0 = λ [ l 3 3 ] 0 2a = λ ( 8a 3 3 ) x y (Δ) −a 0 a δl l δm y (Δ) 0 2a δl l δm x