TS EAPCET (Engineering) Systems of Particles and Rotational Motion — practice questions
38 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice TS EAPCET (Engineering) Systems of Particles and Rotational Motion in the app →The moment of inertia of a solid sphere of mass $M$ and radius $R$ about a diameter is:Two point particles of equal mass are placed at x = 2 m and x = 8 m on the x-axis. Where does the centre of maIf $\hat{i}$, $\hat{j}$, $\hat{k}$ are right-handed orthogonal unit vectors, then $\hat{i} \times \hat{j}$ equFor a homogeneous (uniformly dense) thin circular disc, the centre of mass lies at:The **moment of inertia** of a thin uniform rod of mass $M$ and length $L$ about a perpendicular axis through The **angular momentum** of a particle of momentum $\vec p$ at position $\vec r$ (from origin) is:For a body **rolling without slipping** on a flat surface with centre-of-mass velocity $v$, the **velocity of A solid sphere ($I = \tfrac{2}{5}MR^2$) rolls without slipping down a plane inclined at $\theta$. Its **linearBy **parallel-axis theorem**, the moment of inertia of a thin ring of mass $M$, radius $R$, about a tangent inA uniform rod of length $L$ is suspended from one end. Its **time period** for small oscillations is:A **yo-yo** (solid uniform cylinder) of mass $M$, radius $R$ unwinds under gravity. Its downward acceleration Angular displacement theta in a circular motion is related to arc length s and radius r as:Angular velocity omega in terms of time period T:Linear speed v and angular speed omega relation for circular motion:Torque tau acting on a body produces:Angular momentum L of a body rotating with angular velocity omega:A wheel starting from rest reaches 20 rad/s in 4 s. Angular acceleration alpha:Moment of inertia of a uniform solid sphere of mass M and radius R about an axis through center:Moment of inertia of a thin uniform disc (mass M, radius R) about axis perpendicular to disc through center:Parallel axis theorem states I_parallel = I_cm + Md². Here d is:For a thin rod of mass M and length L, I about axis through center perpendicular to rod:Kinetic energy of a rotating body with moment of inertia I and angular velocity omega:For a body rolling without slipping with velocity v of center of mass:A solid cylinder and a hollow cylinder of same mass and radius roll down an incline. Which reaches bottom firsTwo point masses 3 kg each at the ends of a massless rod of length 4 m. I about axis through center perpendicuA figure skater pulls in her arms during a spin. Her moment of inertia decreases. By conservation of L (I × omFor a hoop, disc, and solid sphere of same mass and radius rolling down an incline. Order of acceleration (fasA solid sphere of mass M, radius R rolls without slipping down an incline of angle theta. Acceleration:Two point masses 3 kg each at the ends of a massless rod of length 4 m. I about axis through center perpendicuFor a thin rod (mass M, length L) about axis through ONE END perpendicular to rod, I equals:Torque on a body produces angular acceleration. For tau = 20 N m and I = 5 kg m², alpha equals:A 50 kg flywheel of radius 0.5 m rotates at 60 rev/min. Angular momentum (I = MR²/2):A solid sphere rolls down without slipping. Fraction of KE that is rotational:A wheel rotating at 50 rad/s decelerates uniformly to rest in 10 s. Total angular displacement:Angular acceleration alpha = 4 rad/s² acts for 5 s on a wheel starting from rest. Number of revolutions:A solid cylinder (I = MR²/2) of mass 2 kg, radius 0.1 m, rolls without slipping at 6 m/s. Total kinetic energyFor a solid cylinder rolling without slipping, the friction force at the contact point:Moment of inertia of an annular disc (inner radius r1, outer r2) about perpendicular axis through center: