{"id":9445,"date":"2026-06-01T21:33:48","date_gmt":"2026-06-01T21:33:48","guid":{"rendered":"https:\/\/kapdec.com\/help\/?p=9445"},"modified":"2026-06-01T21:33:48","modified_gmt":"2026-06-01T21:33:48","slug":"torque-angular-acceleration-and-momentum","status":"publish","type":"post","link":"https:\/\/kapdec.com\/help\/torque-angular-acceleration-and-momentum\/","title":{"rendered":"Torque Angular Acceleration And Momentum"},"content":{"rendered":"

Unit: <\/strong>Torque and Rotational Motion<\/strong><\/h2>\n

Chapter: <\/strong>Torque angular acceleration and angular momentum<\/strong><\/h3>\n

Reference: AP Physics Algebra, <\/em>Torque and Rotational Motion, <\/em>Angular Acceleration and Momentum, Torque, <\/em>Angular acceleration, <\/em>Torque and angular momentum<\/em>, <\/em>Angular Momentum, <\/em>Torque and Rotational Motion, Conservation of Angular Momentum, Equilibrium of a rigid body<\/p>\n

After studying this chapter, you should be able to:<\/strong><\/p>\n

    \n
  • define torque and find the direction of rotation produced by it;<\/li>\n
  • state the principle of angular momentum <\/li>\n
  • state the principle of conservation of angular momentum;<\/li>\n
  • calculate the velocity acquired by a rigid body at the end of its motion on an inclined plane<\/li>\n<\/ul>\n

    Angular acceleration<\/strong><\/p>\n

    Analogous to the kinetic variables of linear displacement (s) and velocity (v) in translational motion, we have angular displacement (q) and angular velocity (w) in rotational motion. It is then natural to define in rotational motion the concept of angular acceleration in analogy with linear acceleration defined as the time rate of change of velocity in translational motion. We define angular acceleration <\/strong>a<\/strong>n <\/strong>as the time rate of change of angular velocity<\/strong>; Thus,<\/p>\n

    \"\"<\/p>\n

    If the axis of rotation is fixed, the direction of w and hence, that of a is fixed. In this case, the vector equation reduces to a scalar equation<\/p>\n

    Torque and angular momentum:<\/strong><\/p>\n

    The turning<\/strong> effect of a force is called torque. <\/strong>Its magnitude is given by<\/p>\n

    τ<\/em>\"\" =Fs =Frsinθ.<\/em><\/p>\n

     <\/p>\n

    Where s <\/em>is the distance between the axis of rotation and the line along which the force is applied.<\/p>\n

    \"\"<\/p>\n

    The units of torque are newton-metre, or Nm. The torque is actually a vector quantity. The vector from of τ<\/em>\"\" =Fs =Fr\/sinθ<\/em>\"\"  is<\/p>\n

    τ<\/em>\"\"  = r <\/strong>× F<\/strong><\/p>\n

    Angular Momentum:<\/strong><\/p>\n

     <\/p>\n

    The product of linear momentum<\/strong> and the distance from the axis is called angular momentum, denoted by <\/strong>L. <\/strong>If we sum this product for all the particles of the body, we get<\/p>\n

     <\/p>\n

    \"\"<\/p>\n

    Remember that the angular velocity is the same for all the particles and the term within brackets is the moment of inertia. Like linear momentum, the angular momentum is also a vector quantity. Eq. (L=Iω<\/em>\"\") gives only the component of the vector L <\/strong>along the axis of rotation. It is important to remember that I <\/em>must refer to the same axis. The unit of angular momentum is kg m2<\/sup> s–1<\/sup><\/p>\n

    Recall now that the rate of change of ω is α<\/em>\"\" and I α<\/em>\"\" = τ<\/em>\"\". Therefore, the rate of change of angular momentum is equal to torque. <\/strong>In vector notation, we write <\/strong>the equation of motion of a rotating body as<\/p>\n

     <\/p>\n

    \"\"<\/p>\n

    \"\"<\/p>\n

    Example 1:<\/strong><\/p>\n

    Shows a bicycle pedal. Suppose your foot is at the top and you are pressing the pedal downwards.<\/p>\n

    \"\"<\/p>\n

     <\/p>\n

    (i<\/em>) What torque do you produce? (ii) Where should your foot be for generating maximum torque?<\/p>\n

     <\/p>\n

    Solution: <\/strong><\/p>\n

     <\/p>\n

    (i) When your foot is at the top, the line of action of the force passes<\/p>\n

    through the centre of the pedal. So, θ<\/em>\"\" = 0, and  τ<\/em>\"\"= Fr <\/em>sin = 0.<\/p>\n

     <\/p>\n

    (ii) To get maximum torque, sin θ<\/em>\"\"  must have its maximum value, that is θ<\/em>\"\" must be 90º. This happens when your foot is at position B and you are pressing the pedal downwards.<\/p>\n

    If there are several torques acting on a body, the net torque is the vector sum of all the torques. Do you see any correspondence between the role of torque in the rotational motion and the role of force in the linear motion? Consider two forces of equal magnitude acting on a body in opposite directions (Fig. given below). Assume that the body is free to rotate about an axis passing through O. The two torques on the body have magnitudes<\/p>\n

     <\/p>\n

    \"\"<\/p>\n

     <\/p>\n

    \"\"<\/p>\n

    We can verify that the turning effect of these torques are in the opposite direction. Therefore, the magnitude of the net turning effect on the body is in the direction<\/p>\n

    of the larger torque, which in this case is τ1\"\" <\/strong>:<\/p>\n

    \"\"<\/p>\n

    We may therefore conclude that two equal and opposite forces having different lines of action are said to form a couple whose torque is equal to the product of one of the forces and the perpendicular distance between them.<\/strong><\/p>\n

    There is another useful expression for torque which clarifies its correspondence with force in linear motion. Consider a rigid body rotating about an axis passing through point O (Fig. given below). Obviously, a particle-like P is rotating about the axis in a circle of radius r<\/em>.<\/p>\n

     <\/p>\n

    \"\"<\/p>\n

     <\/p>\n

    If the circular motion is non–uniform, the particle experiences forces in the radial direction as well as in the tangential direction. The radial force is the centripetal force m ω2<\/sup> r, which keeps the particle in the circular path.<\/p>\n

     <\/p>\n

    The tangential force is required to change the magnitude of v, which at every instant is along the tangent to the circular path. Its magnitude is m a, where a is the tangential acceleration. The<\/strong> radial force does not produce any torque. <\/strong>Do you know why? The tangential force produces a torque of magnitude mar. Since a = rα\"\", where α\"\" is the angular acceleration, the magnitude of the torque is m r2<\/sup> α\"\". If we consider all the particles of the body, we can write<\/p>\n

     <\/p>\n

    \"\"<\/p>\n

    because α is the same for all the particles at a given instant.<\/p>\n

    The similarity between this equation and F <\/strong>= ma <\/em>shows that τ<\/em>\"\" performs the same role in rotational motion as F <\/strong>does in linear motion. A list of corresponding quantities in rotational motion and linear motion is given in the Table given below. With the help of this table, you can write any equation for rotational motion if you know its corresponding equation in linear motion.<\/p>\n

    \"\"<\/p>\n

     <\/p>\n

     <\/p>\n

    With the help of Eq. τ<\/em>\"\" = Iα<\/em>\"\" we can calculate the angular acceleration produced in a body by a given torque.<\/p>\n

     <\/p>\n

    Conservation of angular momentum:<\/strong><\/p>\n

     <\/p>\n

    Equation, <\/p>\n

    \"\"<\/p>\n

     <\/p>\n

    shows that <\/strong>if there is no net torque acting on the body. This means that there is no change in angular momentum, i.e. the angular momentum is constant. This is the principle of conservation of angular momentum. <\/strong><\/p>\n

     <\/p>\n

    Along with the conservation of energy and linear momentum, this is<\/p>\n

    one of the most important principles of physics.<\/p>\n

     <\/p>\n

    The principle of conservation of angular momentum allows us to answer questions such as: How does the direction of the toy umbrella floating in air remain fixed? The trick is to make it rotate and thereby impart it some angular momentum. Once it goes in air, there is no torque acting on it. Its angular momentum is then constant.<\/p>\n

    Since angular momentum is a vector quantity, its constancy implies fixed direction and magnitude. Thus, the direction of the toy umbrella remains fixed while it is in air.<\/p>\n

     <\/p>\n

    In the case of your friend on the rotating stool; when no net torque acts on the stool, the angular momentum of the stool and the person on it must be conserved. When the arms are stretched, she causes the moment of inertia of the system to increase. Eqn. (L=Iω<\/em>\"\") then implies that the angular velocity must decrease. Similarly, when she folds her arms, the moment of inertia of the system decreases. This causes the angular velocity to increase. Note that the change is basically caused by the change in the moment of inertia due to a change in the distance of particles from the axis of rotation.<\/p>\n

     <\/p>\n

    Let us look at a few more examples of the conservation of angular momentum. Suppose we have a spherical ball of mass M <\/em>and radius R<\/em>. The ball is set to rotate by applying a torque on it. The torque is then removed. When there is no external torque, whatever angular momentum the ball has acquired must be conserved. Since the moment of inertia of the ball is (2\/5)MR<\/em>2<\/sup>, its angular momentum is given by<\/p>\n

     <\/p>\n

    \"\"<\/p>\n

     <\/p>\n

     <\/p>\n

    Equilibrium of a rigid body:<\/strong><\/p>\n

     <\/p>\n

    A rigid body is said to be in mechanical equilibrium, if both its linear momentum and angular momentum are not changing with time, or equivalently, the body has neither linear acceleration nor angular acceleration.<\/p>\n

    If the total torque on the rigid body is zero, the total angular momentum of the body does not change with time. Eq. (b) gives the condition for the rotational equilibrium of the body.<\/p>\n

    Example 1:<\/strong><\/p>\n

    Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.<\/p>\n

    Answer <\/strong><\/p>\n

    Let the particle with velocity v be at point P at some instant t. We want<\/p>\n

    to calculate the angular momentum of the particle about an arbitrary point O.<\/p>\n

    The angular momentum is l = r × mv. Its magnitude is mvr sin q, where q is the angle between r and v as shown in Fig. below.<\/p>\n

    \"\"<\/p>\n

     <\/p>\n

     <\/p>\n

    Although the particle changes position with time, the line of direction of v remains the same and hence OM = r sin q. is a constant.<\/p>\n

    Further, the direction of l is perpendicular to the plane of r and v. It is into the page of the figure. This direction does not change with time.<\/p>\n

    Thus, l remains the same in magnitude and direction and is therefore conserved. Is there any external torque on the particle?<\/p>\n

     <\/p>\n

    Key points:<\/strong><\/p>\n

    Angular Acceleration:<\/p>\n

    Angular acceleration is the rate of change of angular velocity with respect to time. It is denoted by the symbol alpha (α) and is measured in radians per second squared (rad\/s2<\/sup>). Angular acceleration can be calculated using the formula α = (ωf<\/sub> – ωi<\/sub>) \/ t, where ωf<\/sub> is the final angular velocity, ωi <\/sub>is the initial angular velocity, and t is the time taken for the change.<\/p>\n

    Momentum:<\/p>\n

    Angular momentum is a measure of the amount of rotational motion of an object. It is denoted by the symbol L and is given by the product of the moment of inertia of the object and its angular velocity. Angular momentum can be calculated using the formula L = Iω, where I is the moment of inertia and ω is the angular velocity.<\/p>\n

    Torque:<\/p>\n

    Torque is the force that causes rotational motion. It is denoted by the symbol τ and is given by the product of the force applied and the distance from the axis of rotation. Torque can be calculated using the formula τ = r x F, where r is the distance from the axis of rotation and F is the force applied.<\/p>\n

     <\/p>\n

      \n
    • In summary, angular acceleration is the rate of change of angular velocity, angular momentum is the measure of the rotational motion of an object, and torque is the force that causes rotational motion. These concepts are fundamental to understanding the behaviour of rotating objects and the laws of rotational motion.<\/li>\n<\/ul>\n
        \n
      • Angular momentum is a measure of the rotational motion of an object or system.<\/li>\n
      • The angular momentum of an object or system is the product of its moment of inertia and its angular velocity.<\/li>\n
      • The moment of inertia is a measure of an object's resistance to rotational motion.<\/li>\n
      • According to the conservation of angular momentum, the total angular momentum of an isolated system remains constant.<\/li>\n
      • An isolated system is one that does not interact with its environment in a way that changes its angular momentum.<\/li>\n
      • The conservation of angular momentum applies to both classical mechanics and quantum mechanics.<\/li>\n
      • If an external torque acts on a system, then its angular momentum can change. However, the total angular momentum of the system and its environment remains constant.<\/li>\n
      • The conservation of angular momentum has many applications in physics, engineering, and technology.<\/li>\n
      • Some examples of applications of the conservation of angular momentum include the study of celestial mechanics, the rotational motion of objects, and the design of gyroscopes and flywheels.<\/li>\n<\/ul>\n

         <\/p>\n","protected":false},"excerpt":{"rendered":"

        Unit: Torque and Rotational Motion Chapter: Torque angular acceleration and angular momentum Reference: AP Physics Algebra, Torque and Rotational Motion, Angular Acceleration and Momentum, Torque, Angular acceleration, Torque and angular momentum, Angular Momentum, Torque and Rotational Motion, Conservation of Angular Momentum, Equilibrium of a rigid body After studying this chapter, you should be able to: […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[622],"tags":[],"class_list":["post-9445","post","type-post","status-publish","format-standard","hentry","category-ap-physics-1"],"_links":{"self":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts\/9445","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/comments?post=9445"}],"version-history":[{"count":0,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts\/9445\/revisions"}],"wp:attachment":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/media?parent=9445"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/categories?post=9445"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/tags?post=9445"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}