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Rolling Objects

Explores equations used to describe the motion of rolling or sliding objects.

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Rolling Objects

When an object is rolling without slipping this means that \begin{align*} v = r \omega \end{align*} and \begin{align*} a = r \alpha \end{align*}. This is also true in the situation of a rope on a pulley that is rotating the pulley without slipping. Using this correspondence between linear and angular speed and acceleration is very useful for solving problems, but is only true if there is no slipping. Also, know that when the object is sliding, kinetic friction is in play. When it is rolling, then static friction. Often, an object will start out rolling and sliding (kinetic friction) until it slows enough that it is rolling without sliding (static friction). One can set up a condition by forcing \begin{align*} v = r \omega \end{align*} and \begin{align*} a = r \alpha \end{align*} in order to find the point where it stops sliding. Finally, when this rolling object rolls down the incline, it gains kinetic energy and loses potential energy, just like any object going down an incline. However, for rolling objects the kinetic energy is split between two forms: rotational and kinetic:

\begin{align*} KE_{Rolling Object} = \frac{1}2 m{v}^2 + \frac{1}2 I{\omega}^2 \end{align*}

Remember, if the object isn't sliding but perfectly rolling: \begin{align*} v = \omega r \end{align*}

Key Equations

\begin{align*}\tau_{net} = \Sigma \tau_i = I \alpha = F_{Friction} R_{RollingRadius} = \mu F_{normal} R_{RollingRadius}\end{align*}

\begin{align*}\text{Static and Kinetic Friction'} \begin{cases} f_s \le \mu_s | F_N| & \text{Opposes potential motion of surfaces in contact}\\ f_k = \mu_k | F_N| & \text{Opposes motion of surfaces in contact} \end{cases}\end{align*}


You throw a bowling ball of mass \begin{align*}m\end{align*} and radius \begin{align*}r\end{align*} with an initial speed \begin{align*}v_o\end{align*} down a flat bowling lane with a coefficient of kinetic friction \begin{align*}\mu_k\end{align*}. Initially, the ball slides down the lane not rotating at all, but after a time \begin{align*}\Delta t\end{align*}, it begins to roll perfectly without sliding. Find \begin{align*}\Delta t\end{align*} in terms of the values given above.

We'll start by drawing an FBD for the bowling ball.

We'll start by applying Newton's second law to the bowling ball. The force of friction is the only force in the x-direction.

\begin{align*} \Sigma F_x&=ma_x && \text{start with Newton's second law}\\ f&=ma && \text{put in friction for the net force}\\ f&=m\frac{\Delta v}{\Delta t} && \text{substitute in the definition of acceleration. } \Delta t \text{ is the time it takes for the ball to begin rolling}\\ f&=m\frac{v_f-v_o}{\Delta t} && \text{substitute in } v_f \text{ and } v_o \text{, where } v_f \text{ is the velocity at the moment the ball begins rolling without sliding}\\ \end{align*}

We now have two unknowns in our equation so we'll use Newton's second law for rotation as our second equation to help us solve this problem. Using Newton's second law for rotation, we're going to determine a value for \begin{align*}v_f\end{align*} in terms of the other values.

\begin{align*} \Sigma\tau&=I\alpha && \text{start Newton's second law for rotation}\\ fr&=I\alpha&& \text{substitute in friction for the net torque}\\ fr&=I\frac{\Delta\omega}{\Delta t} && \text{substitute in the definition of angular acceleration. } \Delta t \text{ is the time it takes for the ball to begin rolling}\\ fr&=I\frac{\omega_f - w_o}{\Delta t} && \text{substitute in } \omega_f \text{ and } \omega_o \text{ for } \Delta\omega\\ fr&=I\frac{\omega_f}{\Delta t} && \omega_o \text{ is zero so we an simplify the equation}\\ fr&=I\frac{v_f}{r\Delta t} && \text{since } \omega_f \text{ is the angular speed when the ball begins rolling without sliding, we can express it in terms of } v_f\\ v_f&=\frac{fr^2\Delta t}{I} && \text{solving for }v_f\\ \end{align*}

Now, we can put that value we just found back into our Newton's second law equation and solve for \begin{align*}\Delta t\end{align*}.

\begin{align*} f&=m\frac{\frac{fr^2\Delta t}{I} - v_o}{\Delta t} && \text{making the substitution}\\ \Delta t&=\frac{mv_o}{\frac{mfr^2}{I} - f} && \text{solving for } \Delta t\\ \Delta t&=\frac{mv_o}{\frac{\mu_km^2gr^2}{I} - \mu_kmg} && \text{substitute the known values for } f \\ \Delta t&=\frac{mv_o}{\frac{\mu_km^2gr^2}{\frac{2}{5}mr^2} - \mu_kmg} && \text{substitute the known value for } I\\ \Delta t&=\frac{2}{3}\frac{v_o}{\mu_kg} && \text{simplify to get the answer}\\ \end{align*}

Interactive Simulation


  1. A solid cylinder of mass, M, and radius R rolls without slipping down an inclined plane that makes an angle \begin{align*} \theta \end{align*}. The cylinder starts from rest at a height H. The inclined plane makes an angle \begin{align*} \theta \end{align*} with the horizontal. Express all solutions in terms of M, \begin{align*} \theta \end{align*} , R, H, and g.
    1. Draw the free body diagram for the cylinder.
    2. Determine the acceleration of the center of mass of the cylinder while it is rolling down the inclined plane.
    3. Determine the minimum coefficient of friction between the cylinder and the inclined plane that is required for the cylinder to roll without slipping.
    4. The coefficient of friction μ is now made less than the value determined in part d so that the cylinder both rotates and slips. How does the translational speed change from above (i.e. larger, smaller, same). Justify your answer.
  2. For a ball rolling without slipping with a radius of \begin{align*}0.10 \;\mathrm{m}\end{align*}, a moment of inertia of \begin{align*}25.0 \;\mathrm{kg}m^2\end{align*}, and a linear velocity of \begin{align*}10.0 \;\mathrm{m/s}\end{align*} calculate the following:
    1. The angular velocity.
    2. The rotational kinetic energy.
    3. The angular momentum.
    4. The torque needed to double its linear velocity in \begin{align*}0.2\end{align*} sec.

Review (Answers)

  1. b. 2gsinθ/3  c. (tan θ) /3 d. The translational, or linear speed increases because some of the energy that would have gone into rotational kinetic energy now goes to linear kinetic energy, hence making the linear speed greater.
  2. a. \begin{align*}100 \;\mathrm{rad/s}\end{align*} b. \begin{align*}1.25 \times 10^5 \;\mathrm{J}\end{align*} c. \begin{align*}2500 \;\mathrm{Js}\end{align*} d. \begin{align*}12,500 \;\mathrm{Nm}\end{align*}

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